Publication of The College of
CALIFORNIA
IRRIGATION PUMPS
Their Selection and Use
C.N.JOHNSTON
::. . : ; :; ■ .
CALIFORNIA AGRICULTURAL
Experiment Station
Extension Service
CIRCULAR 415
IRRIGATION PUMPS
Their Selection and Use
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C.N.JOHNSTON
Suppose You Need to Irrigate
What kind of irrigation pump are you going to buy?
How will you fit your pump to your well?
Do you know how to work out power costs?
What do you need to know about care of pumping equipment?
Ranchers Who Use Irrigation Pumps Know
that although they have come into common use the problems
concerning them have multiplied instead of becoming simpler.
Operators need to understand this specialized equipment so as
to work out for themselves the most efficient and economical
pumping arrangements for their lands.
This Circular
furnishes as complete as possible a discussion of the pumping
units themselves, the special attributes or characteristics of the
several types of pumps, the application of pumps to wells, the
economics of pumping, and the care and maintenance of pump-
ing equipment. The seven equations used in the discussion are
numbered where they appear and printed in heavy type.
The Author:
C. N. Johnston is Professor of Irrigation and Irrigation Engineer in the Experiment
Station, Davis
WHERE TO FIND THE INFORMATION YOU NEED
Several kinds of pumps may be used
Centrifugal
Turbine
Screw-type
Combination or mixed-flow pump runners
Mechanical features of the various pumps
2. Operating behavior of pumps ....
Interrelationships of pump characteristics
Pump performance on characteristic curves
Applications of characteristic-curve relationships
10
Select the pump that meets your needs .
What capacity pump do you need?
Use of a reservoir with a small water supply
Well characteristics
Well-characteristic curves
Specific capacity of wells
Changing water levels in wells
Fitting pumps and wells
Why develop a well?
Submersible pumps
25
4. Costs and methods of installation
How to compute power costs
Installing pumping plants
Other types of pump drive
41
5. Care and maintenance of the pumping plant 51
[3]
A,
lmost one-half of the irrigated acreage in California is served
by means of pumping. Much of the extension of irrigated land will be by this means.
The selection of proper pumping equipment and its design are of great importance to
the prospective buyer, who asks such questions as, "What type of pump should I
buy?" and "Because I need a given amount of water, what size well and pump do I
need?"
While variables not under complete control affect the answers to these questions,
it is possible to describe these variables, show their part in the general problem, and
provide a workable outline for anyone who may have doubts regarding his ability to
make a wise decision in the matter.
The following pages describe the pumps themselves, then take up their performance
characteristics, which are the clues to their adaptability to meet any set of conditions.
The succeeding pages consider briefly the irrigation requirements of crops, the be-
havior of wells, and finally the application of pumps to wells, including typical pump-
ing and power problems and their solutions.
SEVERAL KINDS OF PUMPS MAY BE USED
It is relatively safe to say that little irri-
gation water is being pumped by any-
thing but high rotative-speed equipment
of one sort or another. The few exceptions
are mostly of limited capacity (20 to 50
gpm) in the form of windmills or power-
driven reciprocating pumps, many of
which are only variations of the wind-
driven variety.
The high-speed rotative units are of two
types : One type has a member, or impeller
or runner, which is about the shape of a
ship's propeller. This is a screw type. The
BY WAY OF EXPLANATION
Here are some abbreviations used
frequently in this circular, together
with the terms they indicate,
h — head
Q — flow
t — time
gpm — gallons per minute
cfs — cubic feet per second
rpm — revolutions per minute
h.p. — horsepower
w.h.p. — water horsepower
kw-h. — kilowatt hour
other has a so-called centrifugal impeller.
Its operation differs from that of the screw
type in that the water is thrown out of it
by the centrifugal force resulting from the
high rotative speed of the drive shaft;
hence the name.
Both types supply very large volumes
of water for their size and weight as com-
pared with the reciprocating pumps ; they
are therefore cheaper and more easily in-
stalled than reciprocating pumps of equal
capacity. The centrifugal impeller is the
more common of these two forms and will
be considered first.
The centrifugal pump
Centrifugal force is a by-product of any
rotation and is the tendency of all parti-
cles in the turning mass to continue in
straight-line motion at any instant. The
result of this force is noted when one has
driven a car through mud and then picks
up speed on the road. The mud on the
wheels flies off all around the wheel.
In a pump a shaft must supply the turn-
ing motion. On the shaft are mounted
vanes which radiate outward. They gen-
erally curve as shown in the drawing. The
vanes force the water to turn with the
[4]
shaft and thus generate centrifugal forces
in the water. As a result the water is
thrown outward from the eye or throat
of the impeller to the outer edge of the
vanes, and at the same time it is given a
considerable amount of rotative speed
about the center of the drive shaft. The
combination of these motions given the
water, which are measured as velocities,
stores energy which we measure in one
way as pressure or head. This is the ability
to lift water to some given elevation.
Outside of the impeller there must be a
pump case or housing to retain the water
under control and to lead it into the dis-
charge pipe, which carries it to the point
of discharge. The drawing shows a pump
case about the impeller. In shape it re-
sembles a snail shell. The space around
the impeller increases in cross section in
the direction of rotation, so that in one
revolution it grows from mere clearance
to the full area of the discharge throat.
This type of case is called a volute be-
cause of its spiral shape, and it is designed
to lead the water as efficiently as possible
from the impeller to the discharge outlet.
The drive shaft of a centrifugal pump can
be mounted either vertically or horizon-
tally. Horizontal mounting, as illustrated,
is the more common arrangement.
At times it is desirable to obtain quite
high pressures from a horizontal centrif-
ugal pump. In this case two or more im-
pellers may be mounted on the shaft. The
Split-case double-suction centrifugal pump from
discharge side, with top of case open
to show parts.
discharge from the first is led to the throat
of the succeeding and so on by appro-
priate arrangement of volutes and con-
necting castings in the case. Each impeller
gives added energy to the water so that
the final result is measured in pressure or
head as many times greater than one im-
peller would give as there are impellers on
the shaft.
Usually centrifugal pump impellers
have, in addition to the vanes, two plates
(one on either side of the vane edges)
Single-suction centrifugal with case or volute
partially cut away. A— open space between
vanes; B— vane; C— path of one particle of water
stream passing through pump; D— inlet (or eye)
suction port of impeller; E— outer edge of im-
peller; F— discharge end of volute passage;
G— edge of discharge flange of pump.
[5]
Mixed-flow impeller of closed type.
cast integrally with the vanes. (A single
side plate is occasionally used.) In this
construction a little better control of the
water is achieved than if the vanes alone
are supplied. These plates (called
shrouds) give additional stiffness to the
vanes, which is a structural advantage.
Multiple-runner or multiple-stage
pumps generally come with an even num-
ber of impellers on the shaft. Having an
even number of runners is an advantage
because it is possible to place pairs back-
to-back. The advantage arises from the
fact that the water in turning from the
eye toward the periphery of the impeller
creates a steady thrust on the plate at the
back side of the vanes. With only one im-
peller (or an uneven number) , this thrust
must be resisted at a thrust bearing on
the shaft. With back-to-back impellers
little unbalanced thrust is left; cheaper
thrust bearings are possible, and less
bearing wear may be expected.
Unbalanced thrust in single impellers is
counteracted, particularly where large
capacity is to be provided, by casting a
special pump case in which an impeller
will operate having two opposing throats,
or eyes. In this way the thrust from either
side comes on the common center plate,
and balanced operation is the result.
Such an impeller can provide double
the flow that could be expected from a
similar single impeller having but one
throat and water passage of the same di-
mensions. The volute for a double-suction
pump is exactly like that for a single-
suction unit, except that it is larger so as
to handle the increased capacity.
The turbine pump
A few of the horizontal centrifugal
pumps and most of the vertical ones are
called turbines because they have special
cases that differ materially from the
simple volute case. Turbines may have
one or more stages and are more often
multiple- than single-stage. The case con-
tains stationary vanes that help direct the
water away as efficiently as possible from
the circumference of the runner to the
discharge pipe. These vanes in the case
of the pump mark it as a turbine. Deep-
well turbines frequently have centrifugal
pump-type runners.
The deep-well turbine case is more
compact than the volute case, and that is
why the deep-well turbine lends itself to
use in the well casing, which is limited in
diameter from both mechanical and eco-
nomic considerations. In the deep-well
turbine with two or more stages, the tur-
bine vanes are necessary because without
them the water leaving stage one would
whirl up into stage two, turning with that
impeller so that impeller two would be un-
able to act upon it at all. No work would
be done, and no added pressure would be
achieved above stage one. With the tur-
bine vanes in place, the whirling water
from the first stage is straightened out
smoothly, flowing straight upward, before
it enters the eye of the second stage, which
can now set it spinning and eject it with
added energy.
Screw-type pumps
Some single screw-type pumps are
found in low-lift work delivering water
from streams, ponds, or ditches with a lift
of 2 to 10 or 12 feet. Well turbines rarely
have screw-type impellers. These installa-
tions, as a rule, are multi-stage assemblies.
Usually both shallow and screw-type in-
stallations have vertical shafts.
The screw impeller, like the boat pro-
peller, is simply an inclined plane that
when turned slides the water up over the
plane and thereby lifts it. The inertia of
the water, which tends to keep it in place,
permits the plane to cut under it and lift
it without whirling it too violently.
The low-lift screw is usually single-
stage and is sometimes operated in a
rather crude box-like (even wooden)
housing. The factory product is generally
better designed and better finished than
these crude forms. The chief advantage of
the screw impeller is its capacity to de-
liver more water than the centrifugal type
of impeller for any given size. Perform-
ance characteristics of both types will be
discussed later.
The screw deep-well turbine finds ap-
plication to small-bore wells where a
maximum amount of water is desired. The
stationary turbine vanes are cast into the
cases of these pumps as in the centrifugal
type, and for the same reason.
Combination or mixed-flow
pump runners
If we took the common centrifugal
pump runner and — supposing it were pos-
sible— bent it to give it a cone shape with
the point of the cone cut off at the eye of
the impeller and with the periphery of the
impeller the big end of the cone, we would
have produced a combination or mixed-
flow impeller. Now if the vanes in the
cone-shaped impeller are tilted slightly,
they will slide under the water passing by,
more like the flow through a true screw
impeller.
These changes are accomplished in
varying degree in the design of combina-
tion impellers to get some of the prop-
Deep-well turbine assembly. A— motor; B— pump
head; C— discharge; D— oiler; E— foundation
base; F— well casing; G— column pipe; H— oil
lubing; I— drive shaft; J— bearing; K— spider;
L— upper bowl assembly bearing and seal; M—
bowl; N— suction strainer; O— lower bowl as-
sembly; P— runner or impeller.
erties of both types for field application.
They are used most frequently in deep-
well turbines because they combine the
feature of rather high discharge capacity,
like the screw, with other features that are
favorable to specific applications.
Mechanical features
of various pumps
The centrifugal pump. Earlier illus-
trations have shown variations in hori-
zontal centrifugal pump design. In the
photo the volute case is bolted to the plate
[7]
on the pump pedestal. As a rule the bolt
holes are symmetrical so that it is possible
to rotate the volute case about the pedestal
plate, making the discharge point up,
down, horizontal, or at almost any in-
termediate angle. The intake or suction
opening stays centered on the shaft line,
while the volute is swung on the bolt cir-
cle. This feature affords some flexibility
to the pump when it is being installed.
The drive shaft passes through the back
of the pedestal assembly. It is sealed in a
running fit in a stuffing box on the back
side of the pedestal assembly so air can-
not leak into the pump along the shaft.
One disadvantage, at times, in this form
of pump (known as single-suction) is that
the suction line and discharge line must
be at right angles to each other. This may
present a mechanical problem which
must be anticipated before installation.
The volute case (shown on page 5)
with the double-suction runner is much
more complex than the single-suction de-
sign, for several reasons: (1) The case
must be shaped to permit the passage of
the suction stream around the discharge
volute. (2) The case must be split hori-
zontally to permit introduction of the
drive shaft and double impeller. (3) The
drive shaft must be supported in bearings
at both ends, as well as the two stuffing
boxes required to make a seal on the shaft
when it enters and leaves the split case.
While costly, these are not objections to
this type of pump.
One advantage is that the suction and
discharge ports of the pump are usually
in line so that connecting piping may be
run on a straight line from suction to dis-
charge. This is often quite a desirable
feature at installation time, and it gen-
erally affords a simpler-appearing piping
installation than results when a single-
suction horizontal centrifugal is used.
The true vertical centrifugal may be in
fact a single-suction horizontal centrifu-
gal case and runner set on end with the
suction opening down, and having an ex-
tended shaft to permit power drive some
distance above the pump proper. An
elbow on the discharge of the volute di-
rects the flow upward to the point of dis-
charge. In this mounting the pump case is
quite often submerged some distance
below the water supply surface, and the
inlet may or may not be equipped with a
suction pipe.
GAUGE READS 50 POUNDS PER SQ. IN.
Detail of centrifugal pump used to illustrate effect of
flow on 1,000-foot discharge line with given static lift.
(Single side-suction centrifugal)
[8]
The split-case double-suction centrifu-
gal is occasionally mounted with a ver-
tical-shaft drive similar to the single-
suction installation, but it is not quite so
adaptable to this service as the other type.
Modifications to the suction port on the
case or housing become necessary, and
the unit is both bulky and heavy for some
installations.
The deep-well turbine: Deep-well
turbines as a group are similar in con-
struction regardless of the type of runner.
They are usually driven from the surface
of the ground ; hence the drive shaft has to
be provided from that point to the bottom
of the lowest stage, or bowl.
The water has to be conducted to the
surface of the ground in a discharge pipe
called the column. A 10- to 20-foot suc-
tion pipe with possibly a strainer at the
end is usually provided. The turbine vanes
in the case of each stage bring the water
back to the drive shaft. Therefore the sim-
plest location for the discharge column is
immediately surrounding the drive shaft.
Thus the drive shaft is in the center of the
column pipe, surrounded by the dis-
charged water. Under these conditions the
shaft needs support every 6 to 20 feet,
and these supports — bearings — need lu-
brication.
Rubber is adequately lubricated by
water, and a stainless steel shaft works
well against such bearings in this loca-
tion. The rubber bearings are held in
place by a thin ring threaded inside the
column. The ring has 3 or 4 light arms
running radially down the shaft to a hous-
ing which retains the rubber bearing it-
self. These assemblies are called spiders.
Rubber bearings are safely lubricated
only if wet. If the pump is just being
started, or has stood for a period unused,
it is best to pour water down the shaft to
wet the rubber bearings before starting.
If this is not done injury will occur at the
dry bearings before they are submerged
in the discharged flow through the column
Pipe-
Before rubber bearings were available,
a seamless steel tube was placed around
the drive shaft in order to separate it
completely from the water. This construc-
tion is still frequently used to avoid the
hazard of injury to unlubricated bear-
ings. Bronze bearings are furnished to
center the shaft within this tubing, and a
mechanically satisfactory installation re-
sults if oil bath or other positive lubrica-
tion is supplied.
Automatic oilers serve this oil require-
ment. You need only keep these oiler res-
ervoirs filled regularly. The tubing is sup-
ported within the column pipe in light
guides similar to the rubber bearing sup-
port (spiders) previously described.
The pump head and some form of
power unit are located at the surface of
the ground, which may be as much as sev-
eral hundred feet above the pump bowls.
The head is a heavy cast-iron assembly to
which the column pipe can be suspended
by screwing it in, and the water flowing
from the column pipe can be turned and
discharged horizontally. The assembly
has a point of attachment for the oil tub-
ing over the drive shaft (if oil tubing is
present) and a stuffing box similar to the
centrifugal pump to permit the drive shaft
to pass through it either to the drive
pulley or to a direct-connected motor.
There are no thrust bearings on a deep-
well turbine at the bowl assembly; there-
fore to keep the drive shaft in tension and
straight, and to keep the bottom of the
runners from scraping on the bowls, a
vertical thrust bearing is introduced at or
near the top of the drive shaft but outside
the pump head. For belt-drive pumps a
combination thrust bearing is just below
the pulley — as a rule on the top of the
pump head. Some electric-motor drives
have similar thrust bearings on the pump
head, while others have the thrust bear-
ing mounted on top of the motor. The
motor itself in these installations has a
hollow shaft. Further discussion will
stress the importance of these thrust bear-
ings. The photo shows a turbine assembly
with most of these components.
[9]
OPERATING BEHAVIOR OF PUMPS
These terms are used in
describing pump characteristics
Any given irrigation pump can be operated at varying speed, can lift water various
heights, and can give a fixed rate of discharge for any given combination of the two
variables — speed (revolutions per minute, or rpm) and lift (total head, or head).
That is to say, if the pump is turning at 1,500 rpm and is lifting water 90 feet, it will
give a certain flow. If, now, either the rpm or the lift is changed, a different flow will
result. While these are the bare facts of the situation, before discussing the implica-
tions further we need a basis for understanding such terms as rpm, head, flow, power,
and efficiency.
Rpm. The term revolutions per minute (rpm) needs no further discussion, but this
factor affects some of the others as follows :
(1) The flow, or Q, varies as the rpm varies, or Q oc * rpm
(2) The headf, or h, varies as the square of the rpm, or h cc rpm2
(3) The work done as water lifted varies as the cube of the rpm, or h.p. cc rpm!
The relationships above, called the affinity relations, are important to anyone having
a belt-driven pump because the temptation is always present to change pulley sizes to
change rpm (and thus the discharge from the pump) or to increase the lift it can
operate against. Suppose the rpm before the change in pulley was 1,000 rpm and after
the change became 1,100 rpm. Also suppose the h.p. before the change of pulleys was
20 and that the lift went up per the second above formula. What would be the theoreti-
cal h.p. at the new speed?
20
(1,000) 3
( 1,100) 3 new h.p. requirement
(10)
(ID
10x10x10
llxllxll
1,000
1,331
* The symbol oc means "varies as.
f See further discussion below.
MINUS 5 POUNDS
VALVE CLOSED
DISCHARGE
Illustrating measurement of suc-
tion lift. (Double-suction split-case
centrifugal pump shown in detail)
WATER SURFACE
[10]
Solution: New h.p. requirement = (20 x 1,331) -r- 1,000 = 26.62, which in this case
is an increase of 6.62 h.p. resulting from 100 rpm added speed. In other words, a 10
per cent increase in the speed required 33.1 per cent increase in power under the con-
ditions of this problem. Each set of conditions requires a complete computation, with
the answers probably varying widely from those found in the above solution.
Head or lift. Lift has been defined as head, and it can be measured in units of
pressure. It is therefore necessary to understand how to measure the lift accurately in
order to calculate the pressure. For instance you can measure lift with a rule or tape
of known length. Suppose the measured lift from the center of a centrifugal-pump dis-
charge pipe to a point of delivery was found to be 115.5 feet, a vertical distance. Sup-
pose the discharge pipe is 1,000 feet long and discharges into a ditch as shown in the
drawing on page 8. If the ditch stays full, and if provision is made so that the water
cannot escape from the pump, the discharge pipe will be full. If a pressure gauge is
tapped into the discharge pipe at the pump and located on a level with the center line
of the pipe at the pump, it will read 50 lb. per sq. in. with the pump stopped.
(This is the static pressure or lift.) In this case the following relations exist:
Head = 115.5 feet (as shown on tape) = 50 lb. per sq. in. (as shown on gauge) .
(4) Therefore, 50 lb. per sq. in. = 11 5.5 feet head or 1 lb. per sq. in. = 2.31 feet head
of water.
Now suppose the pump has been started. It is still possible to check the pressure-
gauge reading and measure the net vertical lift from discharge-water surface down to
the center line of the pump discharge. This taped distance is still 115.5 feet, but the
pressure gauge will now read more than 50 lb. per sq. in. Assume the gauge pressure
has become 62 lb. per sq. in. Then 12 lb. per sq. in. has been added to the static pres-
sure, or lift, because of the friction of the water moving through the line. On the basis
of equation 4 one would find 12 lb. per. sq. in. = 12 x 2.31 feet = 27.72 feet head which
has been added to the static lift, making the total discharge lift 115.5 feet + 27.72 feet =
143.22 feet. The pressure gauge is therefore necessary to account for the total dis-
charge lift or head, and the tapeline will not suffice.
You can, with proper training, compute the friction loss in a discharge-pipe line if
the rate of flow and pipe size are known. Adding this friction loss to the measured
static lift gives the total pumping head. This value should check very closely the value
supplied under flow conditions by an accurate pressure gauge, but the method is not
open to everyone, while an accurate pressure gauge can be used universally.
By the same reasoning a vacuum gauge which reads in negative pressure in pounds
per square inch or in feet of head gives more accurate total suction-lift readings than
are possible with a tape line. Negative pressure or negative head or suction lift can be
described as follows : Suppose a centrifugal pump has been operating and the power
is shut off at the same time the discharge valve is closed. The pump case and suction
pipe are full of water at that instant, and if no air leaks into the case through the stuff-
ing box the case will stay full.
The drawing on page 10 illustrates such a pump unit with dimensions given for
further explanation. The distance from the water-supply surface to the center line of
the vacuum gauge is 11.55 feet and, assuming the gauge is sensitive and accurate, it
reads exactly 5 lb. per sq. in. vacuum or 11.55 feet vacuum, depending on how it is
calibrated. The measured column of water (supply surface to center line of vacuum
gauge) is hanging, in effect, from the level of the vacuum-gauge center to the supply
surface and as a result creates negative pressure.
If the pump is started after this check the gauge will show an increase in vacuum
because of friction losses in the suction pipe ; and this new reading, in feet, is the total
[ii]
suction lift to the center of the vacuum gauge. If, as illustrated, the vacuum-gauge
center line is below the center of the pressure gauge, there is an increase of lift equal
to the vertical distance between these two gauges, which must be added to the sum of
the readings of the two gauges to get the gross total lift for the pump. For example,
combining the data for static conditions from the two drawings above, we get:
Discharge pressure Suction vacuum Distance between Total static
gauge gauge gauges lift, feet
Lb./sq. in. Ft. Lb./sq. in. Ft. Assumed in this case*
50 115.5 5 11.55 1.5 128.55
Sometimes the vacuum gauge is above the discharge-pressure gauge. In this case
there is still a difference in elevation, or the vertical distance between them, but it is
subtracted from the sum of the two gauge readings instead of being added. This is so
because the vacuum reading when taken above the discharge-pressure reading over-
laps some of the pressure recorded at the discharge gauge.
For example, suppose the vacuum gauge had been found to be 1.5 feet above the
discharge gauge when checking the data from the two drawings, instead of below as
found in the previous example. Then the total static lift would be 115.5 4- 11.55 - 1.5 =
125.55 feet. Sometimes the vertical distance between vacuum and discharge gauges is
considerable, and the correction for total pumping head is very important.
Flow. Flow or discharge refers to the amount of water being handled by the pump
in a given unit of time. In irrigation the two terms, gallons per minute (gpm) and
cubic feet per second (cfs), are most commonly used. A few other terms such as
miner's inch, acre-foot per hour or day, and acre-inch per hour or day are also at
times referred to. The following supplies the relationship of the terms:
23]
1 gpm = 231 cubic inches per min. (cu. in./min.), or — - cu. in./sec. = 3.85 cu-
in./sec. 60
1 cfs = 1,728 cubic inches per second
1 700
1 cfs is larger than 1 gpm in the ratio of ^-— = 448.83 or 1 cfs = 448.83 gpm
3.85
Since 448.83 is very close to 450, the expression is commonly written 1 cfs = 450
gpm. (It will be used in the following discussion.)
1 southern California miner's inch = — of 1 cfs = just under 9 gpm
1 California statute miner's inch = j? of 1 cfs = a little over 11 gpm
40
1 acre = 43,560 sq. ft.; therefore 1 acre-foot = 43,560 cu. ft.
, . , 43,560 cu. ft. „ ^0rt .
1 acre inch = — = 3,630 cu. ft.
If 1 cfs flows steadily for 1 hour, then 1 x 60 x 60 cu. ft. have been delivered, or
3,600 cu. ft. This value is so close to the 3,630 cu. ft. per acre-inch given above that
one safely says 1 cfs for 1 hour = 1 acre-inch per hour, or 1 cfs for 24 hours = 24 acre-
inches per day, or 2 acre-feet per day.
Power. All of the above terms involve volume and time. Volume fixes weight be-
cause a certain volume of any substance, including water, has a given weight. Power
is the work done per unit of time and is expressed in terms of weight lifted in feet in
a given time. (See box on page 13.)
* This value had to be assumed here but is carefully measured in the field, using a carpenter's
level or the equivalent to transfer levels horizontally so that the vertical difference in elevation
can be scaled.
[12]
1 h.p. = 550* foot-pounds per second = 33,000 foot-pounds per minute
1 h.p. = .746 kilowatt = 746 watts
1 h.p. hour = .746 kw-h. = 746 watts per hour
* Note: 550 foot-pounds per second might be 100 lb. lifted 5.5 feet in 1 second, or 55
pounds lifted 10 feet in 1 second, etc.
Returning to the expression at the start of this paragraph on power, you can under-
stand that, in pumping, foot-pounds of work are accomplished in a given time; then
power may be calculated if you know the weight of water per unit volume, as follows:
1 gallon = 8.33 pounds 1 cu. ft. = 62.4 pounds
From the above and the preceding relationships we derive water horsepower
(w.h.p.) or the work done in unit time in delivering water by the following equations:
cfs x 62.4 x total lift in feet (h) cfs x h
(5) w.h.p. =
(6) w.h.p. =
550 8.814
gpm x 8.33 x lift in feet (h) gpm x h
33,000 = 3,961
Lift in feet in the above is the total pumping head, as measured. In this connection
you should remember in the case of the deep-well turbine that the suction-pipe and
column-pipe friction losses are generally unascertainable. Therefore it is customary to
measure the distance from the pumping water surface in the well to the surface of the
ground, and then add the total lift above the surface of the ground at the well to this
figure, to obtain the apparent total pumping lift. This is a little unfair to the pump but
the only practical way of doing it. The procedure will be discussed later in more detail.
A couple of examples of the use of equations 5 and 6 may make their application
clearer. We can assume the following data have been determined from two pump tests :
Centrifugal pump test
Flow Suction lift Discharge lift Distance between Total lift
(hs) (hd) gauges (h)
gpm feet feet feet feet
975 17.5 72.3 +-2.2 92.0
Deep-well turbine test
Flow Distance pumping water Total lift above Approximated total
to ground surface of ground lift (h)
cfs feet feetf feet
3.2 87.5 18.2 105.7
Computations of water horsepower
r * -f i v, 975 x 8.33 x 92 975x92 _ r_ ,
Centrifugal pump: w.h.p .= — — = = 22.65 w.h.p.
(from equation 5) 33>000 3'961
Turbine pump: w.h.p. = 3.2x62.4x105.7 = 3.2 x 105.7 = ^ ^
(from equation 6) 550 8.814
Any set of test data from a pump test can be used similarly to ascertain the water
horsepower, which is sometimes called the output horsepower.
f In some cases total lift above the surface of the ground is measured with a pressure gauge, in
which case the distance from the center of the gauge to the ground must be added to the gauge
reading. At other times, with short discharge pipes, the lift is checked with a tape by measuring the
vertical distance from ground line to water surface of the discharge.
[13]
There are two other terms that express power, namely brake horsepower and lab-
oratory or test horsepower. These two terms are used by pump manufacturers to ex-
plain the behavior of their pumps. They are of interest to the buyer because they will
be encountered in pump manufacturers' descriptive literature.
Brake horsepower is the power being delivered to the shaft of a pump at the
pulley or flexible coupling. It is the actual effort being supplied by the motive unit
(either electric motor or engine) at that point. It is useful to know this value or series
of values for the performance of a pump, especially for a pump that is to be engine-
driven, but it is not a measure of the amount of power the owner will purchase for an
electric motor-driven unit, for example. The motor is not 100 per cent efficient (See
below for a discussion of efficiency) and the input h.p. or kw-h. per hour to the
motor is what affects the true cost of power to the owner.
It is safer to convert factory data showing brake h.p. to input h.p. by correcting the
power as follows :
brake h.p. n__ . . . , _.,. , , ,
„ . x 100 = input h.p.; and input h.p. x .746 = kw-h. per hour
per cent motor efficiency
input. Later in the discussion, curves showing pump-performance characteristics are
supplied with the input h.p. and input kw-h. per hour shown, but in some instances
i i i i T t iiii iii- i motor efficiency,
not the brake h.p. In these cases, the brake h.p. would be input h.p. x t-(u\
Laboratory or test horsepower is associated principally with turbine pumps.
It is actually the brake h.p. required at the factory for a test of the bowl assembly
alone. It does not include losses resulting from the long drive shafts to be found when
the bowls are put in deep wells. Hence the figures are low and not useful for checking
true input horsepowers.
Efficiency. Everyone is familiar with the term efficiency as the ratio of recovery to
some original effort; in pumping it has a similar meaning. The water horsepower
divided by the input horsepower gives the decimal value of this ratio, and multiplying
this decimal by 100 gives the per cent efficiency. The equation can be written thus :
output h.p. x100 = effid
input h.p.
Suppose in the case of the turbine pump for which an output of 38.37 water horse-
power was computed in the preceding discussion that the input horsepower was 62.0.
Then the efficiency is .0UtPut' ^j.37 x 1Q() = 6lgg x 1QQ = 61 gg ^ 61 g cent
input, 62.0 r
The term over-all efficiency is common in pumping work and particularly for elec-
tric motor-drive units. It includes the losses in efficiency in the motor as well as all the
losses in the pump unit itself. The owner pays for power on the basis of over-all effi-
ciency; hence the term is important in computing or estimating pumping costs.
We can illustrate the derivation of over-all efficiency by a drawing in which the
individual efficiencies of the various parts of the unit are noted. Here, starting with
the suction and strainer with an efficiency of 97 per cent, the bowl assembly with an
efficiency of 81 per cent, the column pipe with 97 per cent, the head of the pump with
98 per cent, and the motor with 91 per cent efficiency, one computes an over-all effi-
ciency of 67.9 per cent by multiplying all of the efficiencies of the component parts
together, or ( .97 x .81 x .97 x .98 x .91 ) x 100 = 67.9 per cent.
When asking for bids on new pumping equipment, specify that over-all efficiencies be
supplied whenever possible.
(This is suggested because laboratory or test efficiencies are likely to be supplied
[14]
MOTOR
PUMP HEAD
> EFFIC. 91
EFFIC. 98
COLUMN PIPE
^
v\/W/W
EFFIC. 97
BOWLS
EFFIC. 81
SUCTION PIPE
> EFFIC. 97
STRAINER
This figure shows roughly how losses occur in such a deep-well pump. For example of the full power
supplied the motor, 91 per cent becomes useful work. Of the energy reaching the pump head only
98 per cent is useful, and so on down to the suction and strainer assembly.
[15]
otherwise, and confusion and possibly error may result from attempting to convert
these data into equivalent over-all efficiency and further into input horsepower.)
Examples of the interrelationship
of the pump characteristics
Often you may have only a portion of the data from a pump test, but because the
characteristics are all interrelated it is possible to derive the missing values. Three ex-
amples follow to illustrate what is meant and to show how partial data can be used to
derive missing portions to complete the picture for any given pump.
Example 1: Assume we know the discharge (Q) = 1.7 cfs; the total head pumped
against, including both suction and discharge lifts, is 97 feet; and the motor is con-
suming 34 horsepower at the meter. It might be desirable to know the over-all effi-
ciency, the water horsepower, and perhaps the net horsepower delivered by the motor
to the pump so that a different type of drive could be substituted if necessary.
Water horsepower is pounds lifted in feet in a fixed time, and using equation 5 we
find the following by substituting the numbers in this problem:
cfs x 62.4 x total lift in feet 1.7x62.4x97 ___ .
550 550 =18.7w.h.p.
From equation 7 by the same process we get
output horsepower 18.7 ~
; x 100 = -r~r x 100 = 55 per cent over-all efficiency.
input horsepower 34
Motor manufacturers supply efficiency curves for their products, and pump vendors
as well as power companies have these data; therefore the efficiency of the motor can
be approximated at this load of 34-horsepower input. Suppose it is found to be 90 per
cent. Then — — - — r— ^ x 100 = 90 per cent = — ^7 — x 100. Therefore output h.p. =
input h.p. ^34 r r
90
Yp-p-x 34 = 30.6 h.p. This is the so-called brake horsepower required by the pump
when lifting 1.7 cfs, a total height of 97 feet. We would have to supply a power unit
capable of delivering this much power continuously to the drive shaft of the pump.
Example 2:* Suppose we wanted to determine how many gallons per minute a
pump would supply if the total pumping head were 105 feet, the over-all efficiency of
the unit were 55 per cent, and the motor could safely use a maximum of 47 horsepower
at the meter. It is possible to ascertain the gpm (Q), the actual water horsepower
(w.h.p.) , and the kw-h. per hour drawn by the motor.
tt • • output h.p. (w.h.p.) _„ __ w.h.p. ,_
Using equation 7 again, *-: — r — x 100 = 55 per cent = — ttt^x 100
& M & input h.p. r 47
, 55x47 __
***-- loo"25-85
By using equation 6 this time (because gpm are involved) we have:
gpm x lift gpm x 105
3,961 =^-bD= 3,961
* Note: This example like several others here is set up to illustrate the method of obtaining cer-
tain operational data by computation, when other information is available as a starting point. Here
the over-all efficiency, the lift, and the actual input horsepower to the meter are known from the
following sources: the over-all efficiency presumably from some pump manufacturer, the lift from
known field data, and the motor-input horsepower from previous experience, or from the manufac-
turer or his representatives. We may often wish to learn with these data available how much water
can be handled, and this problem gives the procedure. Variations in other problems meet different
situations encountered in the field.
[16]
gpm
25.85 x 3,961
105
= 975
The kw-h. per hour used by the motor is 47 x .746 = 35.06 kw-h. every hour.
Example 3: Against what total head can a pump unit with an over-all efficiency of
58 per cent lift 2.2 cfs if the safe motor input is 28 h.p., and what is the water horse-
power?
Using equation 7, as in example 2,
output h.p. (w.h.p.)
input h.p.
58x28
wJup.— jgg-
Substituting in equation 5, we get
cfs x 62.4 x total lift
100 = 58 = w.h.p x 100
16.24 water horsepower
Total lift =
550
16.24 x 550
16.24
2.2 x 62.4 x total lift
550
65.06 feet
2.2 x 62.4
It can be seen that there is a certain order to the solution of these problems, because
the calculations are not taken in order of the listing of the unknowns in the examples.
We have to apply the equation that fits the conditions and have only one unknown left
to get a solution. It is a fact, too, that these computations have been possible because all
of the terms like flow (Q) , lift (h) , efficiency, etc., are interrelated in a given pumping
unit, so that if part of them are available, as a rule, the rest can be calculated. We can
do little if both head and Q are missing from the data, but, with only one lacking,
many other values may be calculated, as well as the missing value of h or Q.
Pump performance
or characteristic curves
How these curves are drawn. The
previous discussion has considered a
given flow or Q from a pump; and the
several related factors such as head, effi-
ciency, water horsepower, input horse-
power, and over-all efficiency have been
provided originally or found by compu-
tation.
Taking example 2 above as typical, we
might decide to make a picture of the
data, both supplied and computed on a
graph. Custom has put the flow, or Q, on
the bottom of the graph and the scales for
the other factors up the left or right side
of the graph (at right). This figure is
finished so far as the data from example
2 are concerned. The method for complet-
ing it in graphical form is as follows and
based on these data: Q = 975 gpm; over-
all efficiency = 55 per cent (minimum) ;
kw-h. used per hour = 35.08; h = 105
feet; input h.p. = 47; water h.p. = 25.85.
25 85 WATER HP
FLOW OR Q IN GPM
Graph of data from one test of a typical pump.
Note there is one horizontal scale (in this case,
gpm). Also note that there are actually 5 vertical
scales, but that in this drawing head and effi-
ciency (over-all) are combined on the left side
and that the three power scales are combined
on the right. Any other combination might have
been worked out for the vertical scales, but Q
would stay at the bottom or be plotted as the
only horizontal factor in the graph.
[17]
A suitable scale in gpm is selected for
the bottom line, placing 975 gpm in this
case about the center of that line. Then to
either the left or right sides of the chart
another group of scales is planned to take
care of the rest of the factors. One scale
generally takes care of input, output
horsepower, or w.h.p., and kw-h. demand.
Sometimes efficiency also uses this scale.
In the graph shown, however, efficiency
and head are on the same scale. This was
purely arbitrary and subject to the draft-
man's decision in order to separate the
points reasonably well. A light dotted line
is run straight up from the 975-gpm point
on the bottom line. Then starting at the
head-and-efficiency scale on the left, hori-
zontal lines are run to the right from 105
feet head and 55 per cent efficiency points
on this scale to intersections with the
dotted line representing 975 gpm.
We have now located two points that
represent respectively the h - Q and the
efficiency - Q characteristics of the pump
for 975 gpm. By the same process except
from right to left the two horsepower
points and the kw-h. point with respect
to Q are spotted on the graph. The graph
is now done. We have to assume that this
flow and the related points plotted are the
result of some constant rpm so that no
variables arise from changing rpm. (See
page 10, equations 1, 2, and 3, which in-
dicate effect of varying rpm.)
If the efficiency of the motor were
known, it would be possible to derive the
brake horsepower also as follows :
, motor efficiency , , ,
input h.p.x — — = brake h. p.
This term could be plotted against Q
along with the other power units already
shown. Brake horsepower would be the
work done by the motor at the pump shaft.
Suppose more data were obtained from
the pump while it was turning at constant
speed and while a discharge valve was
opened and closed so that the total lift
varied considerably. Then many more
test points would be available and other
points could be computed. As the dis-
charge valve was being closed the total
lift would tend to increase and the flow
rate would diminish. Correspondingly,
other points would change up or down in
value, depending on the type of pump
under test. The value of these data lies in
the fact they may also be plotted as in the
graph on page 17 as a series of points for
h - Q, efficiency - Q, etc., that will tend
to form smoothly curved lines represent-
ing the characteristic of the pump over its
full range at that speed.
Pump characteristic curves in gen-
eral: This graph shows a group of pump
characteristic curves. Note that curves
pass through the 975-gpm line at the same
points as those on the first graph. This
was done on purpose to illustrate how
those test data might have been com-
pleted. These curves may duplicate the
performance of some model centrifugal
or deep-well turbine pump at some fixed
rpm, but they cannot be applied to any
specific pump without testing and check-
ing it. In fact, each pump has its own
characteristics, and its curves cannot be
used to study any other unit.
FLOW OR Q IN GPM
Finished graph of characteristic curves resulting
from series of tests on same pump for which one
test was graphed on fig. 10.
[18]
The shapes of these curves are of in-
terest because, if the pump is discharging
any given quantity of water within the Q
scale, a line drawn straight up from that
point on the Q line intersects all the
curves, and the points of intersection
locate the values of h, efficiency, etc., at
which the pump must have been oper-
ating. Thus if the flow becomes quite
small, say 100 gpm, the head must have
become about a maximum, the efficiency
has approached zero, and the other fac-
tors have also changed from the 975 gpm
condition.
Another detail of importance is the fact
that when zero Q is being pumped, or
when the discharge valve is shut off tight
(or if the total lift exceeds 120 feet for
this pump), the pump supplies just 120
feet head and no more. Shutting off the
discharge valve can do no injury to the
pump itself for that reason. Each pump
will have its own shut-off head, which it
will never exceed so long as the rpm stays
constant. Changing the rpm shifts all the
curves somewhat and establishes a new
shut-off head. Direct motor drive pumps
maintain constant rpm conditions.
Over to the right of the chart, above
1,800 gpm flow, the head that the pump
can develop has fallen to 54 feet. This
means that if the total lift is actually 54
feet the pump will be delivering 1,800
gpm. The three power curves, and partic-
ularly kw-h. per hour input and horse-
power input, are not good at all in this
pump for the reason that the minimum
horsepower input corresponds to 600
gpm, while at zero and at 1,500 to 1,800
gpm output the horsepower input has
risen to double or more that required at
600 gpm. If the pumping head varied
widely for this installation, one would
have to supply over 80 horsepower capac-
ity to move 1,800 gpm, 60 horsepower at
close to shut-off capacity, and only about
30 when moving 600 gpm. If an 80-horse-
power motor had to be put on the pump
costly operations would be induced as
well as high initial costs.
One conclusion from the above discus-
sion and from inspection of the graph is
that this pump running at constant speed
has fixed operating characteristics — so
that if the flow is, say, 1,200 gpm, the
total lift must be 97.6 feet and no other,
and that corespondingly water horse-
power, input horsepower, kw-h. per hour,
and over-all efficiency must be fixed as
indicated by the individual curves where
they cross the 1,200 gpm line. There is no
escaping this fact; at any given rpm the
total lift by the pump controls the dis-
charge (Q) , and all the rest of the factors
are then set.
Characteristic curves for centrifu-
gal-type runners: The graph shown be-
low is a set of pump characteristic curves
rather typical of the cenrtifugal-type run-
ners in either horizontal or deep-well tur-
bine cases. A notable difference appears
between the input and kw-h. per hour
power curves on this graph and those on
the previous graph. Here these two power
curves are quite flat. This form is very de-
sirable for the reason that it is impossible
to overload the motor under any condi-
tions, if the motor is sufficiently large to
handle normal pumping loads out near
the peak efficiency.
f 40-
FLOW OR Q IN CFS
Characteristic curves from a pump having
centrifugal-type runners.
[19]
Present h (2 stages),
feet
Q or cfs
Head per stage h.
feet
Present
Per stage
74
71
0
1.0
2.0
3.0
0
1.0
2.0
3.0
37
35.5
31.7
22.5
63.4
45
If we wished to assume that this graph
came from a deep-well turbine having a
centrifugal runner, it is probable there
were two runners or stages in series pro-
ducing the h - Q curve. This statement is
predicated on the actual situation in the
field for nominal sizes of deep-well
pumps, where the head per stage will be
somewhere between 20 and 40 feet. In the
case of this pump, and assuming it is two
stages, then each impeller or stage is de-
livering half the head shown on the h - Q
curve on the basis of the table above.
Should a third stage be supplied with
this pump, the total head would increase
by 3 times the corresponding head per
stage for each discharge rate, or if plotted
on this graph the present h - Q curve
would be shifted upward, at each Q, the
indicated head per stage given in the
table. The power curves would be shifted
correspondingly, but the over-all effici-
ency curve would shift only slightly.
This can be put another way as follows :
Suppose we had available a single, a 2-,
and a 3-stage pump of this design, the
test data shown below would result if we
decided on, say, 3 cfs as the required con-
stant discharge :
For these pumps it is logical from
equation 5 that the power values should
change as a multiple of the number of
runners if the head changes in that ratio,
because in the equation the only thing
that will change will be head. This can be
illustrated as shown at top of page 21.
The significance of these computations
and the preceding table lies in these facts :
(1) To meet increasing total pumping
lifts additional stages can be added to a
given pump. (2) Such an addition to the
pump will in all probability increase
the input horsepower (this will be certain
if the delivery rate is kept the same as it
was before after adding the new runner
or runners). (3) This added horsepower
Pump Characteristics of One-, Two-, and Three-stage Identical
Runners at 3 cfs Discharge
Characteristic
Single stage
Two stage
Three stage
Total pumping head
22.5 feet
54
7.75
14.2
10.25
15
45 feet
54
15.5
28.4
20.5
30
67.5 feet
54*
23.25
42.6
30.75
40 f
Over-all efficiency (per cent)
Water h.p. . .
Input h.p. . . .
kw-h. input per hour
Probable size motor (normal h.p.)
* Note: Sometimes there is a slight improvement in over-all efficiency as the number of stages increases.
f Note: This mathematical discrepancy (45 h.p. seems to be indicated) will be explained later in the
discussion.
[20]
cfs x 62.4 x total lift in feet
550
equation 5
Substituting the cfs and total lift from the preceding table,
3 x 62.4 x 22.5
rr- ' = 7.66 w.h.p. One runner or stage
oov)
3 x 62.4 x 45
550
3 x 62.4 x 67.5
550
= 15.3 w.h.p.
= 22.97 w.h.p.
Two runners or stages
Three runners or stages
input will probably require a new motor
for driving the unit (this is often the
case), and possibly a new and heavier
drive shaft will be needed.
Unfortunately more runners cannot be
added to a pump, short of factory installa-
tion. A complicated machining, balanc-
ing, and precision-fitting job needs to be
done in assembling a multistage pump
of any type, and adding new runners re-
quires factory facilities. The deep-well-
turbine type is the only one actually
suitable for such changes because of the
simpler case design in that unit.
Characteristic curves for screw-
type impellers: As stated earlier the
screw impeller develops a relatively lowT
lift per stage. The graph (at right) gives
the characteristic curves for a typical
multi-stage screw pump. As in the case of
the centrifugal-type pump, the h - Q curve
for the screw pump is the sum of the
heads produced by the individual impel-
lers. Hence it is possible to analyze the
curves for the screw pump in the same
manner as the centrifugal was handled in
Section C above.
Briefly stated, on the assumption that
there are, say, 5 impellers in this screw
pump, the efficiency curve per impeller
would be almost, if not, identical with that
on the graph, while the h - Q and the three
power curves would be one-fifth as high
on the vertical scale as they are at present
for the assumed 5-stage pump.
It has been noted above that the shape
of the power-input curves is of special
interest. These two curves (input h.p. and
kw-h. per hour input) on the graph at
right are typical of screw-type impeller
pumps, which almost all show the tend-
ency to demand more horsepower at or
near zero flow than when pumping a con-
siderably larger flow out in the neighbor-
hood of the peak or high point of the
efficiency curve.
This characteristic has some disadvan-
tages for the owner, which can be illus-
trated by setting up an imaginary field
situation. Suppose this pump were oper-
ating satisfactorily at a discharge rate of 3
cfs. The input h.p. would be about 28, and
a 30-h.p. motor would be doing the job
without any trouble. It is now decided to
plant a lot of small trees and carry water
to them with a tank truck, which will in-
volve the use of about % cfs (90 gpm) for
filling the tank at a reasonable rate. To
deliver just .2 cfs the pump will have to be
discharging against about 73-foot head,
Characteristic curves for a pump with screw-
type impellers.
[21]
and the input h.p. will become 43, the
over-all efficiency about 4 per cent.
The 30-h.p. motor will burn up quickly
under this load, and so the decision to use
the pump for this tank-filling job needs
some practical reconsideration. Follow-
ing the same reasoning, the owner of this
pump always faces the possibility that
someone will partially or completely shut
off the discharge valve — by mistake or
otherwise. If this ever happens while the
pumping unit is in operation, a burned-
out motor will result.
The immediately preceding is an in-
dictment of the screw-type pump only for
the special condition listed (too high
head). As has previously been noted, it
produces a maximum flow with a mini-
mum of pump-case diameter. When the
pumping lift is steady and canot be inter-
fered with and maximum production is
required, it is valuable. The operator who
has such a pump need only remember this
high head-low flow hazard to avoid trou-
ble. The buyer of new equipment that
must meet a wide range of flow rates
should avoid such a pump. The character-
istic curves which can be obtained for any
pump being considered tell the story.
Another difference between the screw
characteristic and the centrifugal pump
characteristic lies in the h-Q curve shape.
Comparing the two h-Q curves on the
graphs on pages 19 and 21, we see that
the screw type tends to be steeper. This
means that for the screw pump a given
change in pumping head will cause less
change in discharge values than a corre-
sponding change in head for a centrifugal
pump — another argument in favor of the
screw pump.
When for one reason or another the
over-all pumping lift changes quite a bit
during the season, the screw-type impel-
ler provides a more nearly constant flow
than the centrifugal type. This is often a
very good argument for the use of the
screw-type impellers in wells, if the shut-
off or near shut-off hazard can be elimi-
nated.
The curves on these last three graphs,
which will be discussed later, can be con-
sidered as coming from rather old units
in which the impellers have worn, consid-
erably reducing the over-all efficiencies.
Either type can be bought now to provide
60 per cent or better over-all efficiency,
but in time these efficiencies tend to be-
come lower.
Characteristic curves of mixed-
flow impellers: Mixed-flow impellers
have been described earlier as a combina-
tion structurally between screw and cen-
trifugal types. Their behavior as indicated
by the characteristic curves is also a com-
bination of the two, with peculiar traits
sometimes added. The graph below shows
the characteristic curves of a combination
or mixed-flow runner pump having h — Q
characteristics at zero flow somewhat like
those given in the two preceding graphs,
which might have been the "parents" of
this mixture. That is, the pump might be a
combination of the other two pumps. As
we noted earlier, this is presumably an
old pump that has worn, and so the effi-
ciency is considerably lower than should
be expected of a new unit.
Several things might appear to come
logically from mixing the characteristics
;_ JNPUTKWK-Q
/
/
V
<c~~
FLOW OR Q IN CFS
Mixed-flow impeller pump characteristic curves.
[22]
of two such different runners: (1) The
head developed per stage for the screw
might be increased. (2) The power curves
might not swing up near zero flow as ex-
cessively as they do in the graph on page
21. (3) The h - Q curve might become a
combination of the h - Q curves for the
two "parents," or at least different from
either. All these things are true, and in
some cases the argument for using the
mixed-flow runners hinges on the benefits
derived as a result of changed characteris-
tics that create most effective performance
patterns. Sometimes the combination de-
sign produces somewhat objectionable
performance features.
The graph at left can be used to discuss
both desirable and objectionable features.
Of the desirable features the relatively flat
input-power curves are of course note-
worthy. The h — Q curve is quite similar to
that for the centrifugal runner pump, ex-
cept for the fact that the maximum head
(highest point on the curve) is some dis-
tance out to the right of zero flow. This
feature is quite often found in mixed-flow
pump curves, and for some to a more
marked degree than on the graph shown.
Such a condition is, in some installations,
a disadvantage, especially in cases where
the h - Q curve bows upward more than it
does in this one.
To illustrate the effect of this bowing,
suppose this pump encountered a total
lift of 73 feet. The h - Q curve rises higher
than that value but drops below it before
zero flow is reached. This means there are
two points on the h - Q curve that have a
head value of 73 feet, and that the pump
can discharge either about .25 cfs or al-
most 1 cfs at that head. The unit will pro-
ceed to do just that, alternating back and
forth in an uncertain manner called
"hunting." Under such flow conditions
the user of the water has considerable
difficulty controlling the discharge
stream, which makes the unit undesirable
at this critical head. It is obvious, if the
upward bow from zero flow to the right is
higher and longer than that shown, that
there is a greater zone of performance
affected by this hunting characteristic.
We have no way of comparing the sizes
of the pumps whose characteristics are
pictured in these last four graphs, but it
is probable that the screw pump is in the
smallest case, the mixed-flow next, and
the centrifugal the largest. This means the
mixed-flow could probably be applied to
a smaller well than the centrifugal type
and produce a desired flow, while the
screw type could be accommodated in a
still smaller casing and produce the same
discharge as the other two. Deep-well tur-
bines are the actual subject of the preced-
ing comparison.
A note of caution should be sounded
here regarding the strict acceptance of
these so-called typical pump curves as
truly typical. Great variations can be
found in the characteristics of a given
type (screw or centrifugal), and when
these two are mixed the results are that
much more diverse. Curves for the cen-
trifugal and screw-type are more "typi-
cal" than that for the last one shown for
the preceding reason. You should not
necessarily use these curves as a basis for
deciding the type of runners responsible
for other sets of characteristic curves. In
fact, the type is not so important as the
shapes of the curves themselves, which
indicate what the pump can do and how
it will accomplish it.
Application of pump
characteristic curve
relationships
It would be a mistake not to note and
be able to put to use the fact that the five
curves — h - Q, over-all efficiency Q, input
h.p. -Q, kw-h. per hour input Q, and
w.h.p. - Q — are all interrelated in one
way or another.
Actually these relationships have been
put to use several times in the previous
discussion, and completely in drawing
the pump characteristic curves. The value
in understanding these relationships lies
in the fact that often only part of the data
23]
, Q x pounds/unit volume x lift in feet
Then, output h.p. = 550 or 33;00o (whichever fits) eq. 5 or 6
are available, but one can build up the output h.p.
rest by starting with what is provided. For input h.p. X 10°- W"h InpUt hP- meaS'
example, given Q and h for a given pump ured, the input kw-h. per hour is input
from equation 5 or 6 (whichever is ap- h.p. x .746.
plicable), you can obtain the water horse- The immediately preceding steps are
power. Usually the input horsepower can the normal path in finishing these data.
be obtained at the electric meter,* while Suppose in contrast the over-all efficiency
the given Q is being delivered. With Q and the input h.p. to be fixed for a
input h.p. available and w.h.p. computed, series of Q values. You can compute the
we obtain over-all efficiency using equa- over-all head against which the pump can
tion 8 where per cent over-all efficiency = operate as follows :
t.. 11 rr- • n \ output h.p. (unknown ) _,...
lirst, over-all efficiency (known) = — : : — ■ x 100 eq. 7
input h.p. (known) n
^ . input h.p. x over-all efficiency „. . ... . _
Output h.p. = — ( Ihis will be a fixed value for each Q
value)
output h.j
_ ... . . output h.p. x 550 or 33,000 (whichever fits)
Or, lift in ieet = ^ , j-t t— — : :
(^ (gpmorcls) x pounds/unit vol.
This is the general form of the equations. Typical solutions may help to make them
clearer.
Problem 1 : Suppose for Q = 1.3 cfs, the over-all efficiency was 52 per cent and the
input h.p. was not to exceed 27.5. We wish to know the maximum over-all head that
this pump could operate against.
- . . , ',_ , input h.p. x over-all efficiency
Solution: w.h.p. or output h.p. = — — — -
27.5 x 52
= —^^- = 14.3 h.p.
Total lift in feet . output hp x 550 (Q is in cfs) = 14^0 = % % f ^
Qx 62.4 (Q is in els) 1.3x62.4
Problem 2: A pump is to be bought to deliver water through an over-all lift of 77
feet, and the motor driving it should not be over 20-h.p. nor use more than that amount,
because the current supply from a local plant was limited. Sixty per cent over-all
efficiency is guaranteed by several pump manufacturers. How many gpm can the
pump deliver?
_ . . , input h.p. x over-all efficiency 20 x 60 , _ .
Solution: w.h.p. =— £ ^ — L = 1QQ = 12 w.h.p.
„ ... output h.p. x 33,000 (Q is in gpm)
Q x 8.33 (Q is in gpm)
nn 12 x 33,000 ^ 12 x 33,000 ^ _ .
77 = Qx8.33 °rQ= 77x8.33 =6l7A^
Problem 3: A pump bought to deliver 875 gpm at a total lift of 146 feet and with
an over-all efficiency of 61 per cent is running with a 40-h.p. motor directly connected
to it. Why does the motor overheat?
* Note: This procedure will be discussed later.
[24]
m . . . , - 875x8.33x146
Solution: From equation 6, w.h.p. = = 32.24 output h.p. or w.h.p.
oo5UUU
. „ . output h.p. x 100 . ,
Over-all efficiency = : L. or input h.p. =
input h.p.
output h.p. x 100 32.24 x 100
over-all efficiency 61
52.85
The reason the 40-h.p. motor was running too hot was that it was overloaded.*
The preceding illustrates what has been stated before. It is possible to develop miss-
ing characteristics of pumps if certain others are available.
* Note: This overload is not so large as it appears on casual inspection. A motor delivers its rated
h.p. or brake h.p. to the shaft at 100 per cent load. If this motor were 91 per cent efficient, it would
40
draw —= 43.96 h.p., or the input h.p. would be 43.96 h.p. at rated capacity. For this reason 52.85
h.p. input is 52.85-43.96 = 8.89 h.p. overload on the input (about 22 per cent overload, which
could cause serious heating) . Such motors can usually be run with 10 per cent overload without
injury, which would mean 44 brake h.p. and 48.36 input h.p. at that overload.
SELECT THE PUMP THAT
MEETS YOUR NEEDS
Consider these factors
When installed, a pump has to meet
certain requirements. A particular dis-
charge rate is wanted, and a given lift
must be overcome. You may wish to have
the discharge rate stay within certain lim-
its while the total lift varies a given
amount. The former situation is more
easily met than the latter, when the
changes in total lift that must be met may
make the limits set for discharge flow
physically impossible to accomplish.
When the desired limits cannot be met, a
compromise has to be arrived at to adapt
the equipment as well as possible to the
actual situation in the field. The curve
that controls the pumping cost is the over-
all efficiency (equation 8).
Keep this in mind when deciding on a
pump to meet your set of conditions: Se-
lect the unit that will meet the situation
while operating at the highest efficiency
or over the highest range of efficiencies.
Earlier discussion has covered the term
total pumping lift, which can be a combi-
nation of suction lift and discharge lift in
the case of the horizontal centrifugal, and
is normally assumed for deep-well tur-
bines and other vertical-shaft pumps to be
the total lift from pumping-water levels
to the surface of the ground, plus the dis-
charge lift above ground surface (See
drawing, page 26) .
For either the horizontal centrifugal or
the deep-well turbine, the pumping-water
levels and the discharge levels may vary
in the field for several reasons. As a re-
sult the total pumping lift can be expected
to fluctuate, probably during the pumping
season or possibly from season to season.
All of the h - Q characteristic curves have
a tendency to show a decrease of Q with
increasing head for all pump types. We
must generally consider this fact and de-
cide that at the maximum expected head
some minimum flow is to be produced by
the proposed pump.
This fixes one point on the h - Q curve,
and the flows produced at the lowest ex-
pected total lift will depend on the shape
of the h - Q curves for the respective
manufacturer's products. Now the over-
all efficiency curves become important.
Perhaps two or three pumps are reason-
[25]
ably close to the desired demands of h — ciency curves may demonstrate that only
Q, so far as meeting the field conditions is one or maybe none is meeting the range
concerned. Inspection of the over-all em- of conditions efficiently.
SOIL SURFACE
MOTOR
s
te
t?
L *-> ~ ' '
A
fvwv
WATER SURFACE
/
§
Horizontal centrifugal and deep-well turbine drawn roughly to compare and contrast methods of
measurement of total lift for each.
A typical problem
Suppose a centrifugal pump is to take suction from a source of water that is subject
to natural changes in elevation, and also that the pipe line it discharges into is opened
at various intervals in irrigating so that the discharge lift changes. Assume the maxi-
mum expected lift would be 77 feet and the minimum 42 feet. Also suppose the oper-
ator of this pump must have at least 500 gpm to irrigate satisfactorily.
[26]
The owner has specified these two limits, 500 gpm at 77 feet total lift, and has given
several manufacturers' representatives this specification as well as the minimum total
lift of 42 feet, requesting specifically that efficiencies and input power data be supplied
as for field conditions (true over-all efficiency or input power values) . The owner then
gets back some tables and some characteristic curves, which he assembles as a group of
500
FLOW OR Q IN GPM
1000
60—
Characteristic curves drawn for 3 pumps, a, b, and c, for comparison.
characteristic curves and a table to see how they meet his problem. The illustration
above gives the completed graph he produced, and the tables that were supplied and
computed for three pumps are shown on page 28.
The curves on the graph indicate that the operator will get his desired flow (500
gpm) when the total lift is 77 feet because all three (pumps, A, B, C) h - Q curves
pass exactly through h = 77 feet at Q = 500 gpm. Flows for the respective pumps will
vary considerably at the minimum head of 42 feet, as shown at the top of page 28.
[27]
Pump
Discharge (gpm)
A
1,040
B
880
C
1,120
This means the discharge from the pumps will change quite a bit from 77-feet head
to 42-feet head, and it appears probable that Pump C would supply him with the
greatest amount of water at the lower head. Consequently the total pumping time might
be shortened and a saving made by using this pump (actually the h-Q curve for
Pump C is above that for both A and B between heads 42 and 77 feet, which means
Pump C can deliver more flow at any head between those two heads) .
This argument might be sufficient to cause the operator to decide on Pump C, but
we can study the situation a little more closely by considering the effect of the over-all
Tabulation of Pump Characteristics As Supplied and As Computed
Characteristic in question
Pump A
Supplied
data
Computed
data
Pump B
Supplied
data
Computed
data
Pump C
Supplied
data
Computed
data
Q in gpm
Head in feet
Over-all efficiency (per
cent)
Output h.p
Input h.p
Input kw-h. per hour . . .
Q in gpm
Head in feet
Over-all efficiency (per
cent)
Output h.p
Input h.p
Input kw-h. per hour . . .
Q in gpm
Head in feet
Over-all efficiency (per
cent)
Output h.p
Input h.p
Input kw-h. per hour . . .
Q in gpm
Head in feet
Over-all efficiency (per
cent)
Output h.p
Input h.p
Input kw-h. per hour . . .
250
81.5
42.0
5.14
12.34
9.2
250
85
33.0
5.36
16.25
12.12
250
77.2
19.50
14.55
25
4.875
500
75
57.5
9.47
16.46
12.28
500
75
55.2
9.47
17.15
12.79
500
75
20.60
15,36
46
9.47
750
66.5
56.1
12.59
22.43
16.74
750
55
56.5
10.42
18.45
13.76
750
71
23.60
17.60
57.0
13.45
1,000
48.0
34.2
12.11
35.4
26.4
1,000
24.5
31.5
1,000
58.4
6.185
19.63
14.65
27.31
20.39
54.0
14.75
[28]
efficiency on the input h.p. or input kw-h. Sometimes it is cheaper in the long run to
pump a little longer at a higher efficiency overall than to pump the shorter time at a
substantially reduced efficiency, because it is the gross consumption of electricity that
affects the power bill. The power bill on a pumping load is considered at length later.
We need to make a few more assumptions (it can be assumed you have made those
that follow before you finally select your pump) before undertaking a study of the
effect of varying discharge rates and over-all efficiencies (both affect h.p. input and
kw-h. per hour use) .
Assume that the pump will have to operate 1,000 hours per year if the minimum
discharge of the pump is delivered continuously. Obviously the minimum flow will
be only a transient situation in the field, but this assumed flow is a starting point. Next,
we can assume that the pump will operate ^4 the time at 77-feet lift, *4 the time at
42-feet lift, and % the time at a lift midway between 77 and 42 feet, or 59.5-feet lift.
This is a rough approximation of what may actually be the distribution of pumping
lifts through the season. The discharges provided by the respective pumps are defined
by these three heads where the h - Q curves cross them. The h - Q data for the three
pumps can be picked off the curves as shown below.
Discharge in Gallons Per Minute
Pump
Head operated against in feet
77.0
59.5
42.0
A
B
C
500
500
500
870
720
990
1,040
880
1,120
The over-all efficiencies and kw-h. per hour input picked from these same curves for
the three pumps operating at these heads are as follows :
Over-all Efficiency and kw-h. Input Respectively
Pump
Head operated against in feet
77
59.5
42
Efficiency
kw-h.
Efficiency
kw-h.
Efficiency
kw-h.
A
57.5
55.2
46
12.28
12.79
15.36
49.0
58.0
55.0
20.4
13.5
20.4
42.0
46.0
40.0
28.7
14.0
21.0
B
C
The total gpm delivered per year on the assumption it takes 1,000 hours of pumping
at 500 gpm will be: 1,000 x 60 x 500 gallons.
Any given pump, it has been assumed, will pump % of a certain period of time at
the low rate, % of that time at some intermediate rate, and % the time at the maxi-
mum rate. We can let this unknown total time = t (hours) . Then, for Pump A, (%t x
60 x 500) + (y2t x 60 x 870) + (%t x 60 x 1,040) = 1,000 x 60 x 500
[29]
Clearing this equation, (*4t x 500) + (V2t x 870) + (%t x 1,040) = 1,000 x 500
or 125t + 435t + 260t = 1,000 x 500
or 820.0t = 500,000
- 500,000 „A, . . . . _
and t = — — — — = 610 hours (total pumping time using rump A)
So %t = 152.5 hours, and %t = 305.0 hours.
By the same procedure for Pump B,
(%t x 60 x 500) + (y2t x 60 x 720) + (%t x 60 x 880) = 1,000 x 60 x 500
t = 710 (total pumping time using Pump B)
!/4t = 177.5, and V2t = 355.0 hours
For Pump C,
(V4tx 60x500) + (%t x 60 x 990) + (%tx 60x1,120) =1,000x60x500
t = 555.5 (total pumping time using Pump C)
%t = 138.87, and y2t = 277.74 hours '
It is evident from the preceding that the three pumps will each require a certain
period of time to pump the required amount of water and that B will consume the
longest time to complete the job. Each will consume a certain amount of power at
each of the three pumping rates ; the calculations for the total power demand follow.
Pump A: High rate
Time used, hours 152.5
kw-h. per hour 28.7
Total kw-h. at each rate 4,376.75
Final total kw-h. used
Pump B: High rate
Time used, hours 177.5
kw-h. per hour 14.0
Total kw-h. at each rate 2,485
Final total kw-h. used
Pump C: High rate
Time used, hours 138.87
kw-h. per hour 21.0
Total kw-h. at each rate 2,916.27
Final total kw-h. used
Returning to the curves on the graph, we find that the pumps A, B, and C will re-
quire different motor sizes. A needs a 40-h.p. motor; B needs a 20-h.p. motor; and
C a 30-h.p. motor.
This situation in itself indicates there may be a preference in Pump B because it is
going to have the lowest first cost, and the preceding calculations indicate this pump
will be cheapest to operate because it uses the least power in doing the job. The power
alone* will cost about 1.25^ per kw-h.; then the power used will cost as follows:
* Each of these pumps will have an additional fixed power cost added to the power-use charges as
follows :
Pump A, 40-h.p., will cost an additional $200 per year for demand charge.
Pump B, 20-h.p., will cost an additional $100 per year for demand charge.
Pump C, 30-h.p., will cost an additional $150 per year for demand charge.
[30]
Intermediate rate
Low rate
Total time
305.0
152.5
610 hr.
20.4
12.28
6,222
1,872.7
12,471.45
Intermediate rate
Low rate
Total time
355.0
177.5
710 hr.
13.5
12.79
4,793.5
2,270.22
9,548.72
Intermediate rate
Low rate
Total time
277.74
138.87
555.5 hr.
20.4
15.36
5,665.9
2,133
10,715.17
Pump Power used, kw-h. Cost each year at 1 .25 1 per kw-h.
A 12,471 $155.89
B 9,549 119.36
C 10,715 133.94
These charges, which are part of the figuring of a power bill, will be taken up in detail
later.
The above data indicate that Pump B will be cheaper to operate than the other two,
so far as power charges are concerned. One factor has been left out in the considera-
tion of costs of operation — that is the labor costs. Pump B runs 100.5 hours longer
than Pump A, and 156.5 hours longer than Pump C. If labor costs $1.00 an hour it
appears that Pump B is still the cheapest. If, on the other hand, labor exceeds $1.00
an hour one of the other pumps may be the cheapest.
There is another (perhaps clearer and simpler) way of figuring the time required
by these three pumps to do the annual irrigating job. If the maximum and minimum
discharge each use % the pumping time and the intermediate flow x/2 the pumping
time, it is possible to set up the following relationships to obtain the average flow for
the whole time.
General relationship :
y^ max. flow + % intermediate flow 4- % min. flow = average flow
For Pump A:
(% x 1,040) + (y2 x 870) + (% x 500) = 820 = average gpm from Pump A
But to do the job, if the flow is 500 gpm, takes* 1,000 hours per the original assump-
tion; therefore
500 x 1,000 = 820 x x (hours) and * = 500x1,000 >_.. ,
— = 610 hours (operating
time for Pump A)
This solution is almost identical to that used previously except that the value t is
left out of the solution in this simpler form. Since the solution closely follows the
previous one and the answer for operating time is found to check for Pump A, it will
not be repeated for the other two pumps.
* Note: From a table located later in this discussion, 500 gpm flowing 1,000 hours would put on
1,111 acre inches. This flow rate, 500 gpm, might have been obtained if the total depth of irrigation
per year and the acreage covered had been given thus: 55 acres irrigated 20.2" deep would use 1,111
acre inches, and 500 gpm would take 1,000 hours to accomplish this from these relationships 500
500
gpm= = 1.111 cfs; 1 cfs. = 1 acre inch per hour. Then to put on 1,111 acre inches at the rate of
1.111 cfs or 1.111 acre inch/hr. would consume ' = 1,000 hours.
What capacity pump basic data in the table on page 33).
do you need? The sun can extract (by evaporation
Under the previous discussion it was and transpiration) about .2 inch depth of
assumed that a certain pumping time per water per day from the soil and crops dur-
season was necessary at a given discharge ing the summer months in most of the
rate. This was based on supplying some agricultural areas of the interior valleys
fixed depth of water to the area being ir- of California. This means that there will
rigated each irrigation season. It is gen- be 30 x .2 or 6 inches of water extracted
eral practice to have between 40 and 80 per month ; this is the amount the irriga-
acres under each cubic foot per second of tion must replace if the crop is to be kept
water supply. The basis for this practice supplied adequately in one of these
rests on the following reasoning (and on valleys.
[31]
One cfs serves 40 to 80 acres. As-
sume that 80 acres in the northern part of
the San Joaquin Valley are being studied
to determine what supply is needed or
what size pump is required. If the evapo-
ration-transpiration from this field is at
the normal rate of .2 inch per day or 6
inches per month, 480 acre-inches will be
required per month to replace these
losses. If 1 cu. ft. per sec. is available, it
will take 480 hours' pumping to supply
the required water. There are 24 hours
per day and 30 days in the month, or a
total of 720 pumping hours available.
480
= %, or only two thirds of the gross
time is apparently used for pumping.
Actually a longer time than the net 480
hours will be required of the pump be-
cause the water will be lost by evapora-
tion and by seepage in the ditches, by
seepage below the root zone in part of
the fields (which is waste) , and by direct
waste often caused by escape from the
tract entirely. It is because these extra re-
quirements are put on the pump to make
up for losses that the practical load per
cubic foot per second supply is assumed
to be 80 acres. There are other minor rea-
sons, such as the desirability of not being
tied to the irrigation work completely
throughout the months and the provision
of time to shift the flow from one field to
another. In reality, to put 80 acres under
each cubic foot per second supply is often
a mistake — 60 acres per cu. ft. per sec. is
the more workable compromise. (Desert
areas often have 1 cu. ft./sec. for 40 to
50 acres, while in coastal areas as many as
100 to 120 acres per cu. ft./sec. are easily
possible.) This 60 acres is an average
figure between the 40 and 80 acres per
cubic foot per second specified earlier for
the interior valleys.
Use a reservoir with
a small water supply
The capacity of pump required for a
given area is in proportion to this general
ratio if the supply is more or less un-
[32
limited. In cases where the supply is lim-
ited by the capacity of its source, an
expedient is possible in the form of a
reservoir. With storage of the pumped
supply, it becomes feasible to pump con-
tinuously at, say, % cubic foot per sec-
ond, or 225 gpm; then to store the flow
for 12 hours in the reservoir; and finally
to draw 1 cfs from the reservoir, while
the pump continues to pump for the other
12 hours each day.
This does not give the 480 hours irriga-
tion time required per month per the
previous solution for an 80-acre farm
with 1 cfs flow, but 12 x 30, or 360 hours.
It does produce all the water possible
from the source and uses it at convenient
times and at higher rates, making pos-
sible satisfactory coverage of the area it
can serve. In this case, probably about
-ttt-x 80 acres could be served, or 60
480
acres maximum under the conditions
specified for the 80-acre area previously
studied. To use this flow for 40 acres or
less would probably be better judgment.
Rates of application of
water for various soil
types and methods
Various irrigation methods are avail-
able for the application of water to crops.
Some of the methods control the rate of
application of water to a degree, and soil
type exerts an effect on all methods of
water application. The following table is
a generalized summary of the available
data on irrigation water application rates
for several main soil types and for four
common types of irrigation. It is not in-
tended to be restrictive in its application,
but is a general guide permitting a con-
siderable departure from the values given
in the table on page 33.
Wide departures are possible from the
values given in the table ; hence it must be
considered a guide and by no means a
strict set of rules to go by. The sun is the
other factor that largely controls the use
or rate of use of the water and thereby de-
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termines how frequently the application
need be replaced.
As previously noted, the sun extracts
about .2" of water per day, by evapora-
tion and transpiration in the summer
months in most of the interior valleys of
the state. Crops having shallow rooting
systems draw upon a limited supply of
stored water compared to those growing
with deep-rooting habits in the same type
of soil. Similarly, a crop like alfalfa,
which roots 6 to 8 feet or more deep, if
grown in sandy soil has much less storage
water to draw upon than if grown in a
heavy loam or a clay, because the sand
cannot store as much water as the clay.
For these reasons, though the sun's ex-
traction may be at a constant rate at any
time and place, the soil storage may vary
widely. Therefore the frequency of irriga-
tion has to be adjusted to the soil type and
crop as well. (For further details, see Ex-
tension Circular 50.)
The applications of the table are as
follows :
Problem 1: Basin irrigation — Sup-
pose it has been decided to block out an
orchard into 1-acre units for irrigation.
Suppose further that the soil is very heavy
clay in one area and light sandy loam in
another, and the grade within the orchard
is not over 1 per cent in either east-west or
north-south directions (between 0-2 per
cent). The lighter soil (sandy loam) will
require a flow of 7.5 cfs per block (per
acre) and the very heavy clay only 2 cfs
per block. The pump supply must be at
least 7.5 cfs for the sandy loam area, and
the stream will have to be divided into
3-plus blocks to irrigate the heavy clay
area, but 7.5 cfs must be supplied.
Problem 2: Suppose coarse, sandy
soil and a clay loam soil were being irri-
gated from the same pump by sprinkling,
and the lengths of sprinkler run happened
to permit the coverage of one acre each.
What size pump would be needed if the
slope on the coarse, sandy soil was be-
tween 0 and 2 per cent, and on the clay
loam between 5 and 8 per cent?
From the table for 0-2 per cent on the
coarse, sandy soil, 2 inches per acre per
hour is permissible, and for the clay loam
with 5 to 8 per cent slope only .15 inch
per acre per hour. If a pump large enough
to supply 2 acre inches per hour to the
coarse, sandy soil were bought, it would
have a capacity of 2 cfs or 900 gpm. The
same pump would supply — as many
sprinklers on the clay loam area or 13%
times as many one-acre runs.
If the pump is to run as efficiently when
serving the clay loam as for the coarse
sand, the owner will have to supply at
least 12 more sprinkler lines to operate
all at once in order to use the 900 gpm
which the pump can safely deliver to the
coarse sand in one run. This is an unusual
situation, but it illustrates the range of
problems that may be met in deciding
what size pump to buy.
Problem 3: Borders and furrows are
in a sense respectively wide and narrow
ditches down which the water is led.
Hence, in the case of the borders, the flow
required depends on how wide they are
in order that the supply can spread over
it evenly. (The table states this in flow,
cfs, per 10 feet of width.) The furrows are
much alike and the flow per furrow can
be stated. Thus if 1 cfs were available one
could have borders 20 feet wide on a
medium silt loam, and they could be 550
to 880 feet long for 0-2 per cent slope.
Furrows on the same soil but with 5-8 per
cent slope would need only .002 cfs each,
and should be only 110 to 220 feet long.
It would take 500 furrows all operating
at once to use 1 cfs. It is common practice
to start the flow down borders and fur-
rows at a high rate, then cut back the
delivery so that a continuous supply is
afforded the length of the strip, and in the
case of borders also the breadth for the
remainder of the irrigation.
It is clear that the size of pump de-
pends, first, on how many acres are to be
served; second, on how much water is to
be put on per acre per irrigation season;
[34]
and third, on some fixed operating period
of days or hours. All this is based on the
assumption that the supply and corre-
spondingly the size or capacity of the
pump are unrestricted.
Unfortunately this is not generally the
case. The well, where this is the source, is
the limiting factor in supply. The pump
must fit the well capacity. By the same
reasoning the well capacity controls the
area that can be adequately irrigated. The
method of irrigation and the soil type
necessitate modifications in the irrigation
program to permit the efficient applica-
tion of water, but the crop requirements
must be met whatever the procedure;
adopted.
Well characteristics
A well is different from a stream or
pond only in the matter of visual evidence
of the supply. In the latter two sources, if
the water is not there it is obvious, and
also if the pump uses the water from a
stream or pond faster than it is supplied
it can be noted by a lowering of the sup-
ply water levels. The same thing goes on
in a well, but it is not so apparent. In fact,
a well has a characteristic h - Q curve for
any given time and, if ground-water levels
are changing, a series of h - Q curves fol-
lows the general change in ground-water
levels for the area as they move.
Well-characteristic curves. The ac-
companying graph illustrates a typical
characteristic curve for a well. Several
things shown on this curve need further
comment. Over at the left margin where
Q = zero the h-Q or drawdown curve
touches 62 feet head. This means when
there is no pumping the water in the well
(static water) is 62 feet down from the
measuring point (presumed to be the sur-
face of the ground) . Out to the right on
the h-Q curve near Q = 2.5 cfs the curve
begins to turn upward, or h begins to in-
crease faster with increasing Q than had
been the case. This means the first under-
ground stratum (maybe the only one) has
begun to be uncovered by the receding
water. In other words, the surface of
entrance to the well (the supply surface)
is being diminished, so that the head in-
creases faster than it had been.
The body of the h-Q curve between
zero Q and about 2.5 cfs is a straight line
up to the point where the first stratum is
being uncovered. In this zone h changes
in direct proportion to the change in Q.
This is a typical h-Q, or drawdown,
curve from an artesian well. It is not a
flowing artesian well, but as the supply
strata are under pressure the top one is
covered a good many feet by standing or
static water.
Almost all wells in California are of
this general type, but the slope of the
curve differs as well as the position of the
static water measurement (h where Q =
zero). Furthermore, the Q scale may be
much different from that given in the
graph. A nonartesian well has no pres-
sure in the water-supplying zone or
zones; hence as soon as water is with-
drawn less and less of the supply area is
available. As a result the h-Q curves
continue an upward trend at an increas-
ing rate throughout their length.
FLOW OR Q IN CFS
A typical artesian well character or drawdown
curve. (First water-bearing stratum down about
depth 120')
[35]
Specific capacity of wells. One not-
able feature of the artesian well h-Q
curve is the straight line section before
the exposure of the top stratum occurs.
It is possible to evaluate the well in terms
of discharge of water per unit of lift
(specific capacity). This can be illu-
strated from the graph as follows :
Head Q
62' 0.0 cfs
104' 2.0 cfs
Difference in h and Q for two points
gives h = 42 feet and Q = 2.0 cfs or 900
2.0
gpm. Specific capacity =—rr = .0476 cfs
4.2
per foot drawdown of
900
42
21.41 gpm
per foot drawdown. Or,
Head Q
80 .86 cfs
114 2.49 cfs Diff. or h and Q = 34
feet for 1.63 cfs, or 734 gpm. Specific
capacity = ■ * = .048 cfs per foot draw-
down (error due to failure to read points
734
on curves accurately) or-
34
21.58 gpm
per foot drawdown. (A slight error in
reading the data from the well character-
istic curve causes this value, 21.58, in-
FlOW OR Q IN CFS
Same drawdown curve as on page 35 (June 1)
plus drawdown expected for Sept. 30 and in-
cluding a pump h-Q curve.
stead of 21.41 gpm per foot drawdown
above.)
The specific capacity of this well in the
straight portion of the curve is about 21
or 22 gpm per foot drawdown, which is
about an average figure. Many wells pro-
duce at higher and others at lower rates.
In general, wells are used so that the pro-
duction (Q) falls safely within the
straight section of these h-Q curves, or
somewhat to the left of the point where
the effect of uncovering the first stratum
is felt.
Changing water levels in wells.
The curve (at left) might characterize
this well on June 1 of some year. By Sep-
tember 30 of the same year the static
water levels might have lowered 10 feet,
which would cause the straight part of
the curve to move upward 10 feet. The
first stratum will still be exposed in this
well at a depth of about 116 feet; thus by
September 30, when the drawdown curve
had shifted upward 10 feet (see graph
below) , it would cross the 116-foot depth
at Q = 2.1 cfs (the drawdown curve for
June 1 had crossed the 116-foot depth at
Q = 2.6 cfs).
The effect of the seasonal lowering of
water tables has been to lower the capac-
ity of the well at this critical point, which
illustrates what has been stated earlier —
that one should require less flow of a well
than the amount that will cause the un-
covering of the top water-bearing stra-
tum. In a sense, the limit of the well is
reached there, and if water tables go down
the Q limit becomes smaller. Fortunately
this uncovering of strata does not enter
into the consideration of most well-char-
acteristic curves because the top stratum
is sufficiently deep to eliminate it as a
factor.
The preceding indicates that a well has
a fixed h-Q curve at a given time and
that a pump also has a fixed h-Q char-
acteristic at all times. It is logical to con-
clude then that, when a pump and a well
are put together, a fixed flow results, be-
cause somewhere an over-all head is ar-
[36]
rived at that is agreeable to both pump
and well and that determines the flow.
This is actually the case. All one needs to
understand is how this "agreeable" head
is reached.
Fitting pumps and wells
The simplest situation. The easiest
situation to meet in putting a pump in a
well, if the well-drawdown curve is
known, is first to have only that curve to
deal with (no changing underground
water levels) and second for the pump to
discharge the water at the surface of the
ground at the top of the well.
Such a simple installation can be as-
sumed using the lower drawdown line
(marked June 1 drawdown on the graph)
and assuming for the present that it is to
remain steady throughout several years.
We need also to assume that the pump dis-
charges into a ditch or the equivalent at
the surface of the ground; therefore the
total lift corresponds to the drawdown
curve head for any discharge rate. If a
given pump h - Q curve is plotted on this
same figure, it will cross the well-char-
acteristic curve at some point. This inter-
section is the discharge of this pump from
this well until conditions change in the
well or until another pump is put in.
Now if we suppose that conditions do
change by September 30 so that the gross
lift in the well is as shown, a new draw-
down line, also shown, represents the be-
havior of the well, and it intersects the
pump h - Q curve at a new point. In this
case, by September 30 the well-water
levels had gone down, which meant the
total lift had gone up. Thus the intersec-
tion with the pump h - Q curve is at a
higher point than for the June 1 condi-
tions. Higher heads on these pumps imply
in general lower flow rates, and this is
true in the graph used. These are rela-
tively simple conditions for a pump to
meet, but so far as the lift in the well is
concerned this figure illustrates the whole
problem.
More complex situations. Pumps
do not always discharge freely at the sur-
face of the ground at the well. Frequently
they discharge into a pipe line of some
sort; and the pressure, or head, they
operate against is affected by the location
of the delivery outlets along the pipe. The
total head they operate against is the dis-
charge head plus the pumping lift in the
well.
The graph on page 38 illustrates a situ-
ation where the water levels in the well
vary during the season (curves b and d)
and where the discharge head above the
top of the well also varies (lowest curve,
a, on the graph). A typical pump h-Q
curve has been drawn for use in noting
the effects on the pump discharge of the
variable over-all head conditions it has to
meet. The lowest head for the pump can
be assumed as the total lift in the well for
any flow and at any season (in other
words along the well-drawdown curves).
Altogether, four intersections occur be-
tween the pump h-Q curve and the two
well-drawdown curves plus the two maxi-
mum lift curves, c and e. These latter two
curves are the result of adding the dis-
charge lift values for any given flow
(bottom curve in the graph) to the corre-
sponding point on the two seasonal draw-
down curves. For example, at Q = 800
gpm the highest discharge head is 10 feet;
and adding this to the spring drawdown
value at 800 gpm (which is 65.5), the
maximum lift for 800 gpm must be 75.5
feet. These two maximum lift curves show
the ultimate head required of the pump
for the two seasons. The discharge pos-
sible at the four intersections with the
pump h-Q curve represents maximum
and minimum discharge rates for each
season ; they are so labeled on the graph.
The situation pictured is typical of
many pumping plants and becomes a part
of the solution when a pump is selected.
It obviously makes the solution more
complicated than the one covered earlier
(page 36), but the procedure is identical
in both. One more complication is often
a part of these problems. Up to this point,
[37]
the figures have shown wells having fixed
spring and fall drawdown curves. In
many areas there is a continuous change
(generally lowering) of water levels, and
so on the graphs the drawdown curves
keep shifting (upward if the total lift in
the well increases) year by year.
Under such circumstances we must
project the probable future drawdown
curves onto the graph on which the pump
characteristics are to be studied, then cor-
rect the drawdown curves for the varia-
tions in discharge lift above the top of the
well (if any). From the corrected curves
we learn the probable maximum pumping
lift in, say, 5 years, and also the mini-
mum, and can, first, specify more closely
what is desired of a pump and, second,
determine whether the pumps described
in the bids meet the requirements, and
which pump is best.
800
FLOW OR Q IN GPM
More complex situation in well and on discharge
line from pump illustrated, to show effect on
pump discharge using given pump h-Q curve.
Curve a is the head developed on the discharge
of the pump with varying flow; Curve b is the
minimum pumping lift in the well (drawdown
curve for highest water levels); Curve c is the
over-all pumping lift (discharge plus lift in well)
for minimum-lift conditions; Curve d is the draw-
down curve for the well for maximum pumping
lift in the well; Curve e is the maximum total lift
for the pump (discharge head plus drawdown).
The discharge head curve can be more
complex than that given on the bottom of
the graph used, where it is a smooth curve
rising from left to right. Some of these
variables can be illustrated if certain as-
sumptions are made as a base.
Problem 1 : Suppose the specific yield
of a well is 30 gpm per foot drawdown
and that this spring the static water is at
32 feet depth. The lowering of water by
fall each year amounts to 12 feet; for the
last few years water tables have gone
down from spring to spring a depth of
4.5 feet and may be presumed to continue
doing so. The pump will discharge into a
pipe line across a flat stretch of land, then
up a hill to another flat or bench. Irriga-
tion on the first flat will be at a low rate
and on the higher bench at a greater rate
(details in curve A on graph on page 39) .
Suppose you want to select a pump that
will be effective and most efficient 5 years
from this coming fall when the minimum
flow is to be 750 gpm. The curves on the
graph were drawn as follows :
Drawdown curve this spring —
Start with static water (Q = zero) where
h = 32 feet, and put a dot there. Then if
the specific yield of the well is 30 gpm
per foot drawdown, for 1,500 gpm de-
1,500
livered the drawdown will be
30
50
feet. Thus the well will draw down to
32 + 50, or 82 feet. Put a point on the
location of h = 82 feet and Q = 1,500 feet,
and connect these two points with a
straight line. This is the curve B drawn
on the graph.
Drawdown curve for spring 5
years later — Water levels can be ex-
pected to recede for the next 5 years at
the rate of 4.5 feet per year. Then 5 years
hence the lift in the well will increase 22.5
feet. Starting at static water this spring at
32 feet, move up the zero Q line (left side
of chart) to 32 + 22.5 = 54.5 feet. Do the
same at Q = 1,500 gpm where h becomes
82 + 22.5 = 104.5 feet and connect the
two points. See curve C on the graph.
Drawdown curve for fall 5 years
[38
later — The spring-to-fall increase in
water depth is expected to be 12 feet. Ac-
cordingly, add 12 feet to static water and
to h at Q = 1,500 gpm for the spring curve
5 years later. Connect the two points, and
the fall curve is drawn as D on the graph.
Maximum lift to flat and to bench
In fall 5 years hence — Add to fall
drawdown curve just drawn the increased
pumping head corresponding to zero Q,
250 gpm, 500 gpm, etc., at those flow
lines; and after connecting the points de-
rive this final broken curved line, E on the
graph.
All of the characteristics of the well and
discharge line are now defined on this
last broken curve, and the pump must
deliver a minimum of 750 gpm at a head
of 115 feet plus, in the fall 5 years hence.
Knowing the general shape of pump h - Q
curves, we might decide to fix on some
point on the "maximum lift to bench in
fall 5 years hence" curve as the limiting
h - Q point for the pump. In this way
more than the minimum Q would be avail-
able at h= 115 feet.
This decision would be subject to spe-
cial consideration in the field. Conse-
quently it will be necessary to assume for
this discussion that the proposed pump,
meeting conditions for delivery to the
bench, is acceptable. In all probability the
assumption is the normal solution be-
cause, if water is to be delivered to the
bench, the over-all head will have to meet
the situation pictured along that portion
of the curve.
You might decide on a Q of 1,000 gpm
at a gross head of 122 feet, or on 1,250
gpm at a gross head of 132 feet. In order
to be specific the former, 1,000 gpm at
122 feet, is accepted for discussion here.
The pump must meet these h - Q condi-
tions in the fall 5 years hence. The pump
vendors would be asked to bid on pumps
to meet these requirements.
Another specification that might be
made is that the over-all efficiency curve
for the pump should be flat in the neigh-
borhood of 750 gpm and both sides of
this figure as far as possible. The reason
for this requirement is to get a pump hav-
ing good efficiency spread over a wide
range of capacities to improve the over-
all economy where such a spread of re-
quirements is expected of the pump. The
pump vendors can select from a number
of models and supply data on one or
more that meet the requirements most
satisfactorily. Then by comparison of the
various pumps offered it is possible to se-
lect the one most suitable for the job. The
demands upon a pump pictured in the
graph (see below) are somewhat unusual
but not seriously so. Almost any conceiv-
able combination of head and Q may be
encountered somewhere so that each in-
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GPM DISCHARGE
Drawdown curves for well and reasonably com-
plicated discharge-head curve above pump
combined to illustrate resulting maximum pump-
ing head. Curve A is the discharge head above
the pump under varying-flow conditions (the
vertical job over Q = 750 gpm is, as explained
in the text, the result of the bench or elevated
portion of the land and not a result of this par-
ticular flow); Curve B is the spring drawdown
curve for the well; Curve C is the drawdown
curve in the spring 5 years from now; Curve D is
drawdown curve for the well in the fall 5 years
from now; Curve E is the maximum pumping lift
(drawdown plus discharge) 5 years from now.
39
stallation of pump to well becomes a spe-
cial problem in itself.
Why develop a well?
Without going into the question of well
development in detail, it must be empha-
sized that no new pump should go into an
undeveloped well. There are several rea-
sons. The first relates to the previous par-
agraphs, in which specific curves were
drawn to picture the well characteristics.
These curves come as the result of mak-
ing a production test on the well. The
production test is not complete until clear
mud-free water is being pumped at (or
considerably above) the maximum ex-
pected draught on the well. The well's
characteristics change until clear water
is being delivered at the maximum
draught.
Actually the completed production test
or development improves the well, so that
the characteristic curves show higher spe-
cific yields than at any earlier stage in the
process. It is almost impossible to fit a
pump accurately to a well that possesses
no characteristic curve. If the mud and
sand are all cleaned out of the well, the
new pump is saved the injury that may
result from handling this material. Some-
times the damage to impellers and bear-
ings caused by developing a well with a
new pump can cut over-all efficiencies in
half. This is obviously a serious situation
to an owner because of immediate injury
to the equipment, and, from the by-prod-
uct, high operation charges from that
day on.
For the previously listed reasons we
recommend that a well test be completed
on any well in which a new pump is being
installed. The test or development of the
well can be made a part of the well-drill-
ing contract. If so handled, the stipula-
tions in the contract should name the
necessity of clearing out all mud and
sand to produce clean, clear water at some
maximum drawndown or depth in the
well, or when some specified discharge
rate has been reached. One or the other
of these limits will have to be given or
left to mutual agreement between con-
tractor and owner so that there is an un-
derstanding on the subject. The well-de-
velopment process has been explained in
greater detail in Exp. Sta. Circular 404.
Submersible pumps
A special adaptation to the deep-well
turbine is the so-called pencil motor,
which can be submerged in water. It has
a very long body and relatively small out-
side diameter and is, therefore, suitable
for installation in wells.
These motors are put at the bottom of
a submersible pump and are small enough
in diameter to permit the flow of water
between them and the well casing or lin-
ing. They are direct-connected to the im-
peller shaft, and since they are imme-
diately below the bowls a relatively short
driveshaft is possible. No oil tubing or
guides are necessary in the discharge col-
umn leading to the pump head. The only
requirement is to run the electrical energy
to the motor in sealed conductors so con-
structed as to prevent power leakage from
them where they pass through the well
water, and also to provide seals at their
junction with the motor-winding ter-
minals.
In some makes of submersible pumps,
an oil line is run down to the motor along
with the electrical conductor. The short
driveshaft is a definite advantage for a
turbine pump in a deep well. The small
diameter of the motor required for well
service introduces for the manufacturer
of these products mechanical problems
that have been difficult to overcome.
Crooked wells that would cause bearing
difficulties for surface-driven pumps can
sometimes be effectively pumped using
the submersible.
[40]
COSTS AND METHODS OF
INSTALLATION
Consider the power schedule
There may appear to be no correlation
between total power used and the cost per
kw-h. which broadly speaking is true, but
there is a relationship between the total
hours of operation, motor size, and the
resulting over-all cost per kw-h. The way
the power-schedule cost table is laid out,
(Continued on page 43)
How to compute power costs
On earlier pages we used some power cost data to compute the cost of pumping
where several pumps were involved and it was necessary to select the most eco-
nomical one. These data came from generally available tables such as the fol-
lowing:
Cost of electric power (revised schedule, 1950)
Agricultural power schedule. Applicable to agricultural power service required
intermittently throughout the year.
Rate:
Size of installation,
h.p. (nominal)
Annual demand
charge per h.p.
Energy charge in addition to demand.
Rate per kw-h. per h.p. per year.
First 1,000 kw-h.
Next 1,000 kw-h.
All over 2,000 kw-h.
1- 4.9
5- 14.9
$6.62
5.56
5.03
4.50
3.97
3.97
1.54c
1.32
1.22
1.11
1.11
1.06
.74c
.74
.74
.74
.74
.74
.53c
.53
15- 49.9
.53
50- 99.9
.48
100-249.9 .
.48
250-499.9 . .
.48
Payments: Demand charge payable in 6 equal monthly installments — May to
October. Energy charge payable monthly as energy is used.
Here are four typical power cost computations
These problems will be solved as demonstrations using this table to get the an-
ver costs for the following situations:
nual power costs for the following situations
Problem
number
Motor name
plate h.p.
Hours operated
per year
kw-h. per year
Remarks
1
2
5
15
40
10
750
1,600
2,520
19,650
62,650
Motor efficiency, 86 per cent
Motor 10 per cent overloaded
3
4
Motor using 44 h.p.
100 per cent load — motor effi-
ciency 89 per cent
[41]
Power bill for problem 1:
5 h.p. motor running with 10 per cent overload will deliver 5.5 h.p. to pump shaft.
5.5 delivered h.p. will require-^- x 100 input h.p. or 6.395 input h.p. Total kw-h. used
for 750 hours = 750 x 6.395 x .746 = 3,578 kw-h.
Power charges from table :
Annual demand charge, $6.62 per h.p. = 5 x 6.62 = $33.10
Energy charge first 1,000 kw-h. per h.p. = 5,000 kw-h. at 1.54^ per kw-h.,
but there are only 3,578 kw-h. total use ; hence 3,578 x $.0154 = 55.10
Total bill, $88.20
Power bill for problem 2:
15 h.p. motor uses 19,650 kw-h. per year.
Power charges from table :
Annual demand charge, $5.03 per h.p. = 15 x $5.03 = $ 75.45
Energy charge first 1,000 kw-h. per h.p. = 15,000 kw-h. at $.0122 = 183.00
Remaining 4,650 kw-h. at $.0074 = 34.41
Total bill, $292.86
Power bill for problem 3:
40-h.p. motor uses 62,650 kw-h. per year
Power charges from table:
Annual demand charge, $5.03 per h.p. or 40 x $5.03 = $201.20
Energy charge first 1,000 kw-h. per h.p. = 40,000 kw-h. at $.0122 488.00
Energy charge next 1,000 kw-h. per h.p. at $.0074 kw-h., but there are
only 22,650 kw-h. left; therefore 22,650 x $.0074 = 167.61
Total bill, $856.81
Power bill for problem 4:
10-h.p. motor operating at 100 per cent load for 2,520 hours per year
Total kw-h. per year = ^- x .746 x 2,520 = 21,122.7 kw-h.
.89
Power charges from table:
Annual demand charge $5.56 per h.p., or 10 x $5.56 = $ 55.60
Energy charge first 1,000 kw-h. per h.p. at $.0132 or 10,000 x .0132 = 132.00
Energy charge next 1,000 kw-h. per h.p. at $.0074 or 10,000 x .0074 = 74.00
Energy charge next 1,000 kw-h. per h.p. at $.0053 or 1,122.7 x .0053 = . 5.95
Total bill, $267.55
It is possible to determine the net cost per kw-h. for each of these four motor-driven
units as follows :
*oo of)
Problem 1 : 3,240 kw-h. cost $88.20 per year or ' = 2.72^ per kw-h.
o,z40
Problem 2: 19,650 kw-h. cost $292.86 per year, or 1.49^ per kw-h.
Problem 3 : 62,650 kw-h. cost $856.81 per year, or 1.367^ per kw-h.
Problem 4: 21,122.7 kw-h. cost $267.55 per year or 1.265^ per kw-h.
[42]
the longer a motor operates the larger the
gross use of power and the cheaper each
kw-h. becomes.
As a result, a small motor running a
long time can do a job that a larger motor
running much less time could also do.
Both might use the same total amount of
kw-h. in completing the job, but the small
motor because it ran a long time would
use power sufficient to move the charges
over into the cheaper part of the table.
We must remember, though, that labor
charges may also accumulate while the
small pump works a longer period to pro-
duce a given irrigation supply, and that
net power savings may be more than
counterbalanced by increased labor
charges. A reservoir used in conjuction
with such a small pumping plant might
bring economies in labor.
This power-schedule study illustrates
one factor in the selection of pumps and
motors that has been omitted until now,
namely, that there is an argument for the
use of as small a pump as possible so that
the power cost can be kept at a minimum.
This argument can be supported by the
probable lower first cost of a smaller unit,
and if a well is involved the smaller pump
will discharge at a lower rate. Then the
actual pumping lifts will be less than for
a larger pumping unit.
These factors all point toward the de-
sirability of keeping the pump sizes as
small as possible. We must keep in mind,
however, that in order to cover a given
area certain capacities are essential to
keeping up with the losses by evaporation
and transpiration, as well as providing for
emergencies and the normal pattern of
operation changes — also, that an opera-
tor has to have some spare time to himself.
The pump size selected is very often a
compromise between all of these factors
and the labor costs.
Reading the watt-hour meter to
check power input: The total power
used is accumulated on the watt-hour
meter, which the power-company meter
readers record at regular intervals. You
should occasionally check the power
being used at the meter by the pump;
there are simple ways of doing this. Each
meter set-up has a "constant" or K, which
is a factor that can be used in making
these checks.
Most meters have a K marked on the
face or on the disc, which is the watt hours
per revolution of the disc (or the power
company can supply the value of K) . The
kw-h. per hour would be arrived at as
shown in the box below.
The table on page 44 gives the disc con-
stants for a group of standard polyphase
watt-hour meters handling 220 volts, 3-
phase current. If the voltage is 440 in-
stead of 220, the K is double that given
in the table. Sometimes extra coupling
transformers are inserted into the control
panel on large pumping units so that
smaller, cheaper meters can be used. In
this case, the K is not applicable; for this
reason it is best to check with the power
company to get the true K for any meter.
kw-h. per hour = Kx
= 3.6 x K x
revolutions counted 3,600 (seconds per hour)
seconds taken for count 1,000 (watts per kilowatt)
rev. counted
seconds taken for count
and h.p. hours = K x
revolutions counted
seconds taken for count
4.83 x K x
3,600 (seconds per hour)
746 (watts per h.p.)
rev. counted
seconds taken by count
[43]
Disc constants, K, of polyphase watt-hour meters 220-volt — 3-phase
(K in watt-hours per revolution)
Name and type of meter
Amperes
5
10
15
25
50
General Electric :
D3
1.25
1.2
2.4
11/3
5/6
2.5
2.4
4.8
2 2/3
12/3
4.0
3.6
7.2
4
2 1/2
6.0
6.0
12.0
6 2/3
4 1/6
12.5
12.0
24.0
13 1/3
8 1/3
D6 and D7
D14
Westinghouse :
C, OA, OB, RO
Sangamo :
H
The heading on the table : Amperes-5, -10,
-etc. refers to the data on the meter-face
plate which identifies its make, type, and
capacity in terms of amperes.
The watt-hour meter K is useful in
checking the performance of a pump
either during a pump test or where it is
desired to see how the over-all efficiency
is being maintained as the pump ages.
Installing pumping plants
Horizontal centrifugals. It is prob-
able that more horizontal-centrifugal
pumps are installed by their owners than
are the deep-well turbine types. Precau-
tions must be taken with both, but they
are different enough to make it desirable
to separate the discussion of the installa-
tion into two parts.
Foundations. Horizontal centrifugals,
being relatively compact, generally come
on a cast-iron base with motor and pump
accurately lined up, so that it is necessary
only to supply a firm level foundation for
the base and to hook up suction and dis-
charge piping. The care taken in these
steps pays the owner by retaining in the
unit the utmost over-all efficiency incor-
porated by the manufacturer. For exam-
ple, the foundation should be firm and
level, as previously noted. Do not put in
the foundation until you have fixed its lo-
cation by careful consideration of two
factors: (1) the relationship of water
supply and position of discharge piping
and (2) the desirability of keeping the
piping plan as simple as possible.
There are two reasons for simple pip-
ing arrangements: (1) simple piping is
generally most economical; and (2) it
usually includes straight runs and smooth
bends, if any are present, both of which
give best results when water is to be trans-
mitted.
It seems, then, that the foundation is
located to suit the piping, which is very
often the case — with the piping in the
form of plans and not coupled up at this
stage. A good heavy concrete casting (See
drawing) will be firm and lasting enough
if sunk well into the ground in stable ma-
terial. The hold-down or anchor bolts are
cast in the foundation when it is poured.
The top of the concrete casting need not
be exactly level, but when the pump base
is set over it the base should be leveled
carefully, using metal shims to hold it at
an exactly level position when it is bolted
down, or to permit grouting around the
whole base with a rich cement mortar.
(This will lock the shims in place and
provide an exactly fitting solid form for
the pump base to rest on.) Under this
procedure, the hold-down bolts from con-
[44]
crete foundation to pump base cannot dis-
tort the latter and throw pump and motor
out of line.
The drawing shows such an installation
and other details of the centrifugal-pump
installation. It is to avoid any disturbance
of the factory alignment of pump and mo-
tor that a good foundation is required,
both at the time of installation and
throughout the life of the pump. A solid,
lasting foundation tends to eliminate vi-
bration. This is mechanically beneficial to
the unit, and unpleasant noises are re-
duced or eliminated from the sound-trans-
mitting system afforded by the piping.
The piping can now be attached to suc-
tion and discharge of the pump, once the
pump base is securely set on the grout, or
bolted securely on the shim plates. In
planning this piping the construction of
the pump had to be considered so that
suction and discharge ports could be ori-
ented correctly when the pump-unit base
was installed on the foundation. Below
PRIMING CONNECTION
WATER
Foundations and piping for two types of horizontal centrifugal pumps.
[45]
you see the pump end of a single-suction
and a split-case double-suction horizontal
centrifugal, with suction and discharge
piping connected.
Suction piping. Starting with the suc-
tion pipe, one of the chief considerations
in the general layout was to provide a
satisfactory suction-pipe arrangement.
This means that the pump was put near
enough to the source so that the suction
line could be as straight and short as pos-
sible. In both the pumps illustrated, an
elbow or 90-degree turn is shown. This is
standard practice, but by heating and
bending the pipe in a smooth curve a
longer-radius bend can be made, which
would have less friction loss.
So-called long-sweep fittings are avail-
able at some pipe yards. These screw or
flange together in assembly and are bet-
ter on the suction line than the standard
elbows. Straight, short suction lines pro-
vide as low suction lifts as possible. This
is of great importance in centrifugal-
pump installations. They will lift water
20 feet or more on the suction side, but
they will operate much better if the gross
suction lift can be kept below 10 feet.
On the bottom of each of the suction
pipes illustrated on page 45 is a check
valve. This is a flap valve that opens when
water is drawn through it to the pump but
closes when the tendency for flow re-
verses. This feature is made part of these
installations because, by holding the wa-
ter in the system, check valves, or foot
valves, keep the pumps full of water
(primed) , and it is possible to start them
at any time so long as they stay full of
water. In the sketch a priming connection
is shown at the top of both pump cases.
With a foot valve installed, it is neces-
sary only to fill the suction line and pump
with water, pouring it through the prim-
ing connection to prime the pump. The
discharge line has to be closed off or else
rise above the pump case in order to ac-
complish the result. Where no foot valve
is provided the priming connection can
be used as a suction inlet from a vacuum
pump of some sort, if the discharge pipe
has a shut-off valve handy to the pump.
In fact, that valve becomes necessary
if no foot valve is installed. Assuming the
discharge valve is closed and a common
hand-pitcher pump suction is threaded
into the priming connection, the pump
case and suction can be partially evacu-
ated of air. Water from the source will
then follow the reduced internal pressure
area to the pitcher pump, priming the
centrifugal so that its motor may be
started.
Leaks. If there are leaks in the pump
suction or discharge, priming the pump
by pumping out the air may be difficult or
impossible, depending upon how great
the leaks may be. Leaks at piping joints,
if present, have to be taken up by tighten-
ing them further, or by applying heavy
paint, grease, or other mastic material
along the leaking joints. Leaks in the
pump occur at the stuffing boxes, or seals,
on the drive shaft. These stuffing boxes
have a seal called packing inside them.
The packing is held tightly by a sleeve and
hold-down bolts.
With a new smooth pump shaft and re-
silient packing, the hold-down bolts or
nuts need be only hand-tight to assure
good air seals at these points. When it
becomes necessary to force the packing
down with greater pressure, there is dan-
ger of producing excessive friction on the
pump shaft, which will cause scoring of
the metal. When scoring begins a vicious
circle has started, because it has become
more difficult to hold a seal with the pack-
ing, more pressure is needed, and more
scoring results.
Discharge piping. The discharge
pipe from a centrifugal pump is some-
times smaller than the suction. This is so
because the manufacturers wish to keep
the suction lifts as small as possible, and
a larger suction pipe promotes lower suc-
tion lifts. The suction pipe should be no
smaller than the pump provides at the
suction-inlet connection. The discharge
pipe from the pump has less effect on its
[46]
operation than has the suction pipe on
excessive suction lift. Therefore long-
sweep bends are not fundamental. Dis-
charge valves are needed if control of the
discharge rate or other reasons dictate.
The slip- joint coupling shown on the
split-case centrifugal discharge is manda-
tory when the steel discharge pipe is ter-
minated in a concrete standpipe or simi-
lar relatively weak structure, as shown.
Without the slip joint the temperature
changes natural in the pipe will cause ex-
pansion stresses at the stand that may
fracture it, causing leaks. The only other
solution for such installations — a difficult
one — is to make some sort of stuffing box
in the stand within which the steel pipe
can move without creating leakage.
One can observe from the drawing on
page 45 that the split-case double-suction
pump lines up suction and discharge
nicely, but that the single-suction pump
has the suction offset from the discharge.
As we saw before, this latter type of pump
may have the pump case rotated around
the impeller by unbolting the case from
the back plate bolts and rebolting it in the
position that puts the discharge in satis-
factory alignment. Sometimes in so doing
the setting, or foundation, has to be
switched around. Thus, in planning the
arrangements of piping and foundations
this fact must be given consideration. Oc-
casionally it is possible to set these pumps
so that the suction looks toward the sup-
ply and a curved pipe can be run from
suction connection to source, providing a
smooth, free-flowing entrance for the wa-
ter. Curved pipes cannot be threaded into
a pump directly but require a union or
flanges at the pump.
Deep-well turbines. These, as we
have seen, are relatively much more bulky
than centrifugal pumps with their long
column pipes, drive shafts, and so forth.
The installation is, therefore, a special job
requiring heavy lifting equipment, large
clamps, heavy pipe tongs, etc. For this
reason, the seller usually supplies outside
help to make the installation.
Several precautions are taken in such
installations. All joints are threaded, and
great care is taken that all threads are
wiped clean from dirt of any sort before
they are lubricated with red lead or its
equivalent and turned down in place.
Equal care is taken that threads start ac-
curately matched so that no cross thread-
ing is permitted to injure the starting
turns. Every joint is set up tight, and the
final adjustments of tubing and column
length are made at the pump head where
they terminate.
The pump head has suspended on it all
the weight of everything below. Often the
top of the casing is used as the supporting
surface for the pump head. This is gen-
erally satisfactory if the casing is cut
smoothly and at right angles to the casing
length. Any tendency for the pump head
to tip up or hold up from the squarely cut
casing should be corrected by shimming
so that no bending strain is permitted. If
the casing itself has an angle off the verti-
cal, this means the pump head will and
should stand at this same angle, because
in so standing it indicates a straight line
from head through shaft — at least for
some distance.
If heavy steel beams are provided for
the pump head — or better yet a concrete
foundation — similar precautions to those
taken with the casing should be followed.
The pump head should line up with the
casing. These independent foundations
are safer than use of the well casing as a
support for the pump head. Heavy
wooden timbers are sometimes used, but
the tendency is to install and forget them,
and time causes deterioration or shifting,
creating misalignments that go unnoticed
and unattended until mechanical failure
results. They are not recommended.
Turbine discharge pipes. The dis-
charge pipe from a turbine might dupli-
cate that for the horizontal split-case
centrifugal on page 45. The valve is
optional. The slip joint is important when
the discharge enters a concrete standpipe
and, if that is the case, it might be very
[47]
desirable to supply a discharge flap valve
on the end of the discharge pipe inside the
stand.
Where a deep-well turbine discharges
into a long line below the water surface,
a large amount of water could run back
down through the pump unless a flap valve
stopped it at the time the pump was shut
off. Water running down the pump
column tends to turn the pump back-
wards, and this rotation can unscrew the
drive shaft. The result of unscrewing the
drive shaft is generally mechanically dis-
astrous. Severe stresses result that may
rupture the pump head, oil tubing, or
column pipe — sometimes all three.
For this reason, a flap valve at the
stand is necessary where back flow is pos-
sible. On a long, closed steel line, a check
valve near the pump is quite important.
A few pumps have nonreversing features
built into the unit, but a check or flap
valve provides a worthwhile safety factor
for these units too.
Deep-well pump adjustments. The
pump, after installation, must be "ad-
justed" so that the runners turn freely
without touching either top or bottom of
the bowls. This adjustment is of necessity
accessible to the operator at the ground
surface. It comprises a threaded section
of the top end of the drive shaft fitted with
two hex nuts supported by a heavy-duty
ball bearing with a thrust plate.
This ball bearing, mounted firmly in
the pump base or on the motor head, as
the case may be, supports the complete
weight of the drive shaft and impellers
when the bottom nut is screwed down the
shaft far enough to put the whole shafting
assembly in tension down to the bottom
bowl. If this bottom nut is screwed fur-
ther down the drive shaft, the shaft is
raised through the bearing until the run-
ners touch the top of the bowl cases. By
releasing the nut, the runners drop with
the shaft until they seat on the bottom of
the bowl case. The "adjustment" is at
some point between top and bottom posi-
tions for the impellers in the bowls.
In reality the runners as a rule perform
best when they turn freely as close as pos-
sible to the bowl seats, or when they are
as low as they can be made to turn freely
when in operation. This is often deter-
mined by trial-and-error procedures in
the field. It is best to have the "adjust-
ment" of the runners made by an expert
and not attempt it on one's own initiative.
A check of the adjustment is afforded by
the watt-hour meter which will record a
greater use of electrical power if the run-
ners are rubbing than when they are free.
These adjustments are especially im-
portant for runners having the lower side
open, or unshrouded. With this type, wear
on the exposed vanes can be compensated
for by lowering the runners a few thou-
sandths of an inch, and efficiencies often
respond by an increase of 10 to 20 per
cent overall. Closed impellers having two
shrouds do not benefit particularly by the
critical shaft adjustments just described.
Submersible pumps obviously cannot be
"adjusted" while in the well.
Other types of pump drive
Direct motor-driven electric pumps are
most common, but other forms of drive
and sources of power are available. Belts
are one of the most common forms of
transmission used for driving pumps. The
horizontal centrifugal generally has a
horizontal pulley. Therefore a straight
running belt (no twist) needs to be used
from any horizontal shaft-power source
(including motors, fuel engines, and
power take-offs from tractors) . Deep-well
turbines are vertical, so that the pulley
mounted on the drive shaft stands verti-
cally. As a result, horizontal-shaft power
units have to be connected with these pul-
leys by putting a 90-degree twist in the
belt.
Sometimes enclosed geared boxes are
mounted at the top of the pump head with
a pulley shaft parallel with the ground.
These permit coupling internal-combus-
tion engines or other power units to them
directly through flexible or universal
[48]
couplings. It is possible to drive such
geared heads with a straight belt from a
horizontal-shaft power unit. The coupling
between the fuel engines and geared heads
should be resilient to take up the shock of
the power strokes of this type of drive
unit.
Certain advantages are inherent to belt-
drive units or to fuel-engine driven units.
The revolutions per minute of the pump
can be changed relatively simply by
changing pulley sizes in the first type, and
by simple throttle adjustments for the
second. The problem of overloading dis-
cussed under equation 3, page 10, is al-
ways a possibility where increased speed
is given the pump. Thus it is well to watch
the driving unit carefully for a while after
such a change has been made. Overheat-
ing is the sign of such overloading in
most power units.
Tables 1, 2, and 3 give the horsepower
capacities of flat and vee belts. The veloc-
ity of a belt in feet per minute is computed
as follows : Feet per minute = rpm of drive
pulley x diameter of drive pulley in feet x
3.1416. For example, a 6-inch-diameter
pulley is turning 1,740 rpm. How fast is it
driving the belt?
Solution: 1,740 x % x 3.1416 = 2,733
feet per minute
The second table, for quarter-turn or
crossed belts, shows the space necessary
between drive and driven pulleys. It is
essential to provide this spacing for
crossed belts so that they do not rub and
so that the transition from the horizontal
to the vertical pulleys can be made gradu-
ally. Under best conditions, the spacing of
two horizontal pulleys should be about the
same as that for the crossed belts. The rea-
son for the longer spacing shown for
engine drive is the shock effect contrib-
uted by this source. Vee belts can be run
straight or crossed.
Table 3 shows the capacities of five
sizes of vee belts — through E — and the
belt speeds possible. The vee belt clings to
the pulley and therefore does not have to
be set up so tightly as the flat belt, which
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[49]
Table 2. Belt Sizes for Quarter-turn Drives
Horsepower
Leather belts
Rubber belts
Minimum center distance
transmitted
Width
Ply
Width
Ply
Motor drive
Engine drive
10
inches
4
5
6
8
10
12
16
number
1
2
2
2
2
2
2
inches
5
5
6
8
10
12
16
number
3
4
4
5
5
5
6
feet
10
12
14
16
18
20
24
feet
15
15
18
20
22
30
25
50
30
75
34
125
37
will slip unless reasonably tight. This slip-
page causes a waste of power. Some slip is
almost inherent with flat-belt drives.
Fuel engines including automobile en-
gines are used for pumping. Only the
heaviest-duty engines of this type can
stand continuous full-load operation.
Automobile engines cannot. To supply 40
horsepower from a car engine one should
have one capable of giving 80 horsepower
intermittently (this is the way they are
rated for intermittent service) .
Electric motor sizes and efficien-
cies: Motors can be bought in only cer-
tain stock sizes, as listed in table 4. The
approximate efficiencies at full load or
rated load are also detailed in the table.
The motor rating is the horsepower it will
Table 3. Horsepower Transmitted by Horizontal V-belt Drives
Belt speed, feet per minute
Horsepower per belt
A
B
c
D
E
1,000
0.9
1.7
2.4
2.8
1.2
2.3
3.2
4.2
3.0
5.5
7.5
9.0
5.5
10.0
14.5
17.5
7.5
2,000
14.0
3,000
19.5
4,000
23.5
Table 4. Approximate Efficiencies of Induction Motors at Rated Load
Motor size, horsepower
Efficiency,
per cent
Motor size, horsepower
Efficiency,
per cent
5
86
88
89
89
89
90
30
90
7.5
40
91
10
50
91
15
20 . .
60
75
91
91
25.
100
91
[50]
deliver on the drive shaft. Thus if the
motor is delivering 45 horsepower at 90
per cent efficiency it is actually drawing
45/.90 = 50 h.p.
Most motors can be put in service to
deliver up to 10 per cent overload, which
means to deliver 11 h.p. from a 10-h.p.
motor, 22 from a 20-h.p. motor, and so on.
For this reason, from the table, a 40-h.p.
motor could be loaded to 44 h.p.; but if
45 or 46 h.p. were the load, then a 50-h.p.
motor should be used. Overloading causes
heating, and heat ruins insulation in most
motors, shortening their lives. For this
reason it is not recommended that any
overload be put on them.
CARE AND MAINTENANCE OF
THE PUMPING PLANT
Since a pumping plant is relatively ex-
pensive, the effort to keep it in best con-
dition is justified. Care and maintenance
are immediately in order once a satisfac-
tory initial installation has been com-
pleted. Housing comes first to eliminate
dust and dirt and to keep out precipita-
tion.
The size of the pump house is deter-
mined by the equipment, it being neces-
sary only to provide working space all
around it. Three feet clearance is gen-
erally enough. When the pump is running,
ventilation is required to take off the heat
developed by the power unit. Therefore
a door and windows that can be opened
are generally installed. Sometimes louvres
are put in the sides of the pump house to
permit continuous ventilation, with cross
draft if possible.
In cramped quarters, servicing any
equipment is difficult. When the operator
has to wedge himself into too cramped an
area to perform a regular servicing task,
it may seem easier just to forget the job.
Then lubrication or some other attention
is omitted until injury results to the in-
stallation.
All-weather motors are available that
require no housing. However, adequate
housing would appear to offer some
benefit to such units by shading and pro-
tection from dirt. Both pump and motor
bearings need oiling, and the oil con-
tainers provided by the manufacturers
should be filled regularly so that the sup-
ply never runs out.
As the pumping plant ages, check the
performance occasionally to be sure over-
all efficiencies have not dropped so low
that power costs have become excessive.
In addition, a certain amount of clean-up
is desirable on the equipment, both for
appearances and (more important) to
eliminate the possibility of dirt working
into the moving parts.
Keep stuffing boxes tight but at the
same time draw down the take-up nuts as
lightly as possible. After several thousand
hours' operation, repacking of the stuffing
boxes may be necessary. Sometimes the
addition of a new ring of packing is all
that is needed. It is safer, if the old pack-
ing has become hard or grit has been im-
bedded in it, to remove the old and put in
new. Packing has a lubricating material
in it and is obtainable through the pump
supply houses in the locality.
Most power companies willingly check
the performance of electric-driven pumps
on their system, and it is generally pos-
sible for any customer to obtain this serv-
ice. These operators with power-company
test service, and others who are not so
supplied, may wish to check well-water
levels occasionally to see what the under-
ground water supply situation is. The
drawing on page 52 illustrates two
methods that may be used in "sounding"
these wells to determine static-water and
pumping-water depths. The two methods
are, respectively, an air line and a measur-
ing tape or line.
In the drawing the measuring line is an
[51]
insulated wire which completes a circuit
with a battery and a meter of some sort
in series with it. One can substitute a
magneto for the battery in the wire
sounder and have a bell ring when contact
with the water is made. A tape line or a
string with a small disc or can on the end
may sometimes be used to sound a well if
there is enough room for them to pass
-PUMP
T
down between the column pipe and the
casing. The tape line is run down so that
the end is well immersed in the water.
Then the net dry length is noted to get the
depth to water. The can or disc can be
lowered to the water surface, and it is
generally possible to feel them make con-
tact. If they submerge, the net dry length
can be measured as with the tape line.
ELECTRICAL CURRENT FLOW INDICATOR
BATTERY
AIR LINE
Va" OR >/»" PIPE
AIR LINE
OPERATION
DISTANCE BOTTOM OF AIR LINE
TO TOP OF CASING = b + a -f c
(c CAN BE MEASURED)
PRESSURE GAUGE READING IN
POUNDS PER SQ. IN. X 2.31 = SUB-
MERSION = b
a + b-fc, LESS COMPUTED b AND
LESS MEASURED c, GIVES a, THE
DISTANCE TO WATER FROM THE
GOUND SURFACE
CONTACT ON CASING
WELL CASING
ELECTRICAL SOUNDING LINE
INSULATION
WATER SURFACE
Jk
BARE WIRE
LEVEL IN WELL
u
DETAIL OF END
OF SOUNDER
ELECTRICAL CONTACT
OPERATION
EQUIPMENT
1— BATTERY
2— AMMETER OR VOLTMETER
3— GROUND CONTACT
4— SOUNDING LINE
LOWER SOUNDING LINE TILL RECORD-
ING INSTRUMENT SHOWS FLOW OF CUR-
RENT BY MOVEMENT OF NEEDLE. READ
DISTANCE ON LINE TO WATER FROM TOP
OF CASING (a i c).
SUBTRACT c FROM THIS READING. GIV-
ING a. DEPTH TO WATER
Devices for measuring depth to water.
[52]
Tapes fail for this purpose when water
is leaking into the casing above the water
surface in the well. The falling water wets
the tape and obscures the wetted mark
due to submersion. It is to preclude effec-
tive continuous contacts with falling water
on the electrical sounding wire that the
terminal is guarded as suggested here.
Actually, apparent contacts with the
water surface will result from introduc-
tion of the terminal into a continuous
falling stream, giving surface indications
that otherwise would mean the water sur-
face had been reached. However, when
the line is moved up or down, the indi-
cator will show erratic readings in these
falling streams. The readings suddenly
become steady when the true water sur-
face is reached.
The air line is, as shown in the sketch, a
small pipe run down some distance below
the lowest water level in the well. It has
an open end at the bottom, and, if a pump
is connected to the closed upper end, it
can produce sufficient pressure in the pipe
to force the water out. Any extra air put
into the pipe above that pressure will pass
out the bottom of the pipe. The pressure
read in pounds per square inch (from
formula 4) can be converted to feet of
submergence of the pipe. All one needs to
know, then, is the total length of the pipe
to calculate the depth to water, thus:
Total length of air line (soil surface to
bottom of pipe generally) minus submer-
sion = depth to water.
Some pressure gauges can be set to
read directly the depth to water in feet,
by providing reversed scales that can be
set to read the length of the air line when
the needle is at zero pressure. Air lines
and tapes cannot be used in some wells
because there is no clearance for them,
because the well is too crooked, or be-
cause the pump head provides no inlet for
this purpose. Under these conditions, effi-
ciency tests of the pumping unit are im-
possible either for the individual or for
anyone else.
On new, deep-well pumping-plant in-
stallations the owner can provide a means
for measuring the depth to water by pro-
viding a pipe, as shown in the drawing
(below), that leads from the ground or
foundation elevation to and through the
side of the casing. By making the edges of
the pipe ends smooth so that the measur-
ing line will not be scratched or injured,
and by providing a screw cap for the top
of the pipe, entrance to the well and the
water levels is always possible.
None of these difficulties beset the
normal horizontal-centrifugal setting be-
cause accurate vacuum- and discharge-
pressure gauges can easily be connected
to them and total lifts can readily be ob-
tained.
One of the indications that water levels
are getting too low for the pump is the
discharge of air, or intermittent discharge
from the pump. When this stage is
reached with a deep-well turbine, the
pump must be lowered 20 to 40 feet to
put it down in the supply deep enough to
provide safe continuous operation. This
step may involve reworking the bowls at
the factory and perhaps the installation
of a new motor and drive shafting.
Failing discharge from a horizontal
centrifugal with air in the discharged
water may be the result of leakage of air
through the stuffing box. This is more apt
to occur when the suction lifts are exces-
sive.
i it /o°,o "O \'
CAP
FOUNDATION
PIPE FOR SOUNDING LINE
"-WELL CASING
Small pipe with capped cover provides special
entrance for sounding line in wells otherwise
impossible to sound.
[53]
If water appears in the container on a time passes can throw excessive strain
deep-well turbine-drive shaft oiler, there upon pump castings, causing distortion
is reason to believe the oil tubing has cor- or failure by cracking off parts of the
roded through or has failed in some way. affected structure. Repairs to obvious
Bearings will not last long in a pump so changes in alignment of parts resulting
affected. from these natural movements may save
Settlement in foundations or piping as many dollars.
In order that the information in our publications may be more intelligible, it is sometimes necessary
to use trade names of products and equipment rather than complicated descriptive or chemical
identifications. In so doing, it is unavoidable in some cases that similar products which are on the
market under other trade names may not be cited. No endorsement of named products is intended
nor is criticism implied of similar products which are not mentioned.
20m-8,'52(9869)AA
[54]
IT JUST COULD BE • . .
that the farm problems troubling
you have also troubled others.
And it's also possible that with
a little help from the right source
your problems can be eased, if
not cured.
Here's how to go about getting
help.
Take your problems to your
County Farm Advisor. He's an
agricultural specialist with a
background of practical knowl-
edge about farming in your lo-
cality. He will help you if he can
... or he will get the information
you need from someone who does
know the answers.
Ask your Farm Advisor for a
copy of AGRICULTURAL PUBLI-
CATIONS—a catalog that lists
the bulletins and circulars pro-
duced by the University of Cali-
fornia College of Agriculture, or
write to the address below.
You'll be amazed at the wide
range of information covered in
these publications.
Yes ... it just could be that
your problems aren't nearly as
hard to solve as you think. Make
use of the free services of your
University.
Office of Agricultural Publications
22 Giannini Hall
University of California
Berkeley 4, California